Chapter 8. Work, Energy, and Power
Work
81. What is the work done by a force of 20 N acting through a parallel distance of 8 m? What force will do the same work through a distance of 4 m?
Work = (20 N)(8 m) = 160 J ; F (4 m) = 160 J; F = 40.0 N
82. A worker lifts a 40 lb weight through a height of 10 ft. How many meters can a 10kg block be lifted by the same amount of work?
Work = (20 lb)(10 ft) = 200 ft lb;
Work = Fs = mgs; ; s = 2.77 m
83. A tugboat exerts a constant force of 4000 N on a ship, moving it a distance of 15 m. What work is done?
Work = (4000 N)(15 m); Work = 60,000 J
84. A 5kg hammer is lifted to a height of 3 m. What is the minimum required work?
Work = Fs = (5 kg)(9.8 m/s^{2})(3 m); Work = 147 J
85. A push of 30 lb is applied along the handle of a lawn mower producing a horizontal displacement of 40 ft. If the handle makes an angle of 30^{0} with the ground, what work was done by the 30lb force?
Work = (F cos )s = (30 lb) cos 30^{0} (40 ft)
Work = 1040 ft lb
8
8 N
6. The trunk in Fig. 810 is dragged a horizontal distance of 24 m by a rope that makes an angle with the floor. If the rope tension is 8 N, what works are done for the following angles: 0^{0}, 30^{0}, 60^{0}, 90^{0}?
Work = (F cos )s = (8 N) cos 0^{0} (24 m) = 192 J
Work = (8 N) cos 30^{0} (24 m) = 166 J ; Work_{60} = 96 J ; Work_{90} = 0 J
87. A horizontal force pushes a 10kg sled along a driveway for a distance of 40 m. If the coefficient of sliding friction is 0.2, what work is done by the friction force?
Work = (F cos )s = (F) (cos 180^{0})s =  F s; but F = _{k}N = _{k} mg
Work = _{k}mg s = (0.2)(10 kg)(9.8 m/s^{2})(40 m); Work = –784 J
*88. A sled is dragged a distance of 12.0 m by a rope under constant tension of 140 N. The task requires 1200 J of work. What angle does the rope make with the ground?
Work = (F cos )s;
cos = 0.714; = 44.4^{0}
Resultant Work
89. An average force of 40 N compresses a coiled spring a distance of 6 cm. What is the work done by the 40N force? What work done by the spring? What is the resultant work? Work_{40} = (40 N)(0.06 m) = 2.40 J, (positive work)
Work_{sp}_{ }= (40 N)(0.06 m) = 2.40 J, (negative work)
Resultant work = (works) = 2.4 J – 2.4 J = 0 J
Work is positive when force is with displacement, negative when against displacement.
810. A horizontal force of 20 N drags a small sled 42 m across the ice at constant speed. Find the work done by the pulling force and by the friction force. What is the resultant force?
Work_{40} = (20 N)(24 m) = 2.40 J, (positive work)
Work_{sp}_{ }= (20 N)(24 m) = 2.40 J, (negative work)
Resultant force and, hence, resultant work are zero.
*811. A 10kg block is dragged 20 m by a parallel force of 26 N. If _{k} = 0.2, what is the resultant work and what acceleration results.
F = _{k}N = _{k}mg F = 0.2(10 kg)(9.8 m/s^{2}) = 19.6 N
Work_{ }= F_{R} s = (P – F)s; Work = (26 N – 19.6 N)(20 m) Work = 128 J
F_{R }= (26 N – 19.6 N) = 6.40 N; ; a = 0.640 m/s^{2}
*812. A rope making an angle of 35^{0} drags a 10kg toolbox a horizontal distance of 20 m. The tension in the rope is 60 N and the constant friction force is 30 N. What work is done by the rope? What work is done by friction? What is the resultant work?
(Work)_{rope} = (60 N) cos 35^{0} (20 m); (Work)_{r} = 983 J
(Work)_{F} = (30 N)(20 m) = 600 J; (Work)_{F} = 600 J
Resultant Work = (works) = 983 J – 600 J; Resultant Work = 383 J
Extra work can show that for this example, _{k} = 0.472
*813. For the example described in Problem 812, what is the coefficient of friction between the toolbox and the floor. (Refer to figure and information given in previous problem.)
F_{y} = 0; N + (60 N) sin 35^{0} – (10 kg)(9.8 m/s^{2}) = 0 ; and N = 63.6 N
_{k} = 0.472
*814. A 40kg sled is pulled horizontally for 500 m where _{k} = 0.2. If the resultant work is 50 kJ, what was the parallel pulling force?
F = _{k}N = _{k} mg = 0.2(40 kg)(9.8 m/s^{2}); F = 78.4 N
(P – F) s = 50 kJ; (P – 78.4 N)(500 m) = 50,000 J; P = 178 N
*815. Assume that m = 8 kg in Fig. 811 and _{k} = 0. What minimum work is required by the force P to reach the top of the inclined plane? What work is required to lift the 8 kg block vertically to the same height?
Minimum work is for P = W sin 40^{0} with zero acceleration.
; W = mg = 78.4 N
Work_{P} = P s =(W sin 40^{0}) s = (78.4 N) sin 40^{0} (18.67 m); Work_{P} = 941 J
(Work)_{V} = W h = (78.4 N)(12 m); (Work)_{V} = 941 J
*816. What is the minimum work by the force P to move the 8kg block to the top of the incline if _{k} = 0.4. Compare this with the work to lift it vertically to the same height.
F_{y} = 0; N = mg cos 40^{0}; N = (78.4 N) cos 40^{0} = 60.06 N
F = _{k}N = (0.4)(60.06 N); F = 24.0 N
F_{x} = 0; P – F – mg sin 40^{0} = 0; P = F + W sin 40^{0}
P = 24.0 N + (78.4 N) sin 40^{0}; P = 74.4 N; Recall that s = 18.67 m from Prob. 815.
Work_{P} = (74.4 N)(18.67 m) ; Work_{P} = 1390 J
From Prob. 815, the work to lift vertically is: (Work)_{V} = 941 J
*817. What is the resultant work when the 8kg block slides from the top to the bottom of the incline in Fig. 811. Assume that _{k} = 0.4.
The resultant work is the work of the resultant force:
F_{R} = mg sin 40^{0} – F = (78.4 N)sin 40^{0} – 0.4(78.4 N)cos 40^{0}
F_{R} = 26.4 N; Work = (26.4 N)(18.67 m) = 492 J
Work and Kinetic Energy
818. What is the kinetic energy of a 6g bullet at the instant its speed is 190 m/s? What is the kinetic energy of a 1200kg car traveling at 80 km/h? (80 km/h = 22.2 m/s)
E_{k} = ½mv^{2} = ½(0.006 kg)(190 m/s)^{2}; E_{k} = 217 J
E_{k} = ½mv^{2} = ½(1200 kg)(22.2 m/s)^{2}; E_{k }= 296 kJ
819. What is the kinetic energy of a 2400lb automobile when its speed is 55 mi/h? What is the kinetic energy of a 9lb ball when its speed is 40 ft/s? (55 mi/h = 80.7 ft/s)
E_{k} = ½mv^{2} = ½(75 slugs)(80.7 ft/s)^{2}; E_{k} = 244,000 ft lb
E_{k} = ½mv^{2} = ½(0.281 slugs)(40 ft/s)^{2}; E_{k }= 225 ft lb
820. What is the change in kinetic energy when a 50g ball hits the pavement with a velocity of 16 m/s and rebounds with a velocity of 10 m/s?
Consider the upward direction as positive, then v_{o} = 10 m/s and v_{f }= 16 m/s.
E_{k} = ½mv_{f }^{2} ½mv_{o}^{2} = ½(0.05 kg)(10 m/s)^{2}  ½(0.05 kg)(16 m/s)^{2}
E_{k} = 2.50 J – 6.40 J = 3.90 J ; The change represents a loss of kinetic energy.
*821. A runaway, 400kg wagon enters a cornfield with a velocity of 12 m/s and eventually comes to rest. What work was done on the wagon?
Work = ½mv_{f}^{2}  ½mv_{o}^{2} =(0)  ½(400 kg)(12 m/s)^{2}; Work = 28.8 kJ
*822. A 2400lb car increases its speed from 30 mi/h to 60 mi/h? What resultant work was required? What is the equivalent work in joules?
v_{o} = 30 mi/h = 44 ft/s; v_{f} = 60 mi/h = 88 ft/s;
Work = ½mv_{f}^{2}  ½mv_{o}^{2} = ½(75 slugs)(88 ft/s)^{2}  ½(75 slugs)(44 ft/s)^{2};
Work = 217,800 ft lb ; Work = = 295 kJ
*823. A 0.6kg hammer head is moving at 30 m/s just before striking the head of a spike. Find the initial kinetic energy. What work can be done by the hammer head?
E_{k} = ½mv^{2} = ½(0.6 kg)(30 m/s)^{2}; E_{k} = 270 J
Work = E_{k} = 0 – 200 J; Work = 270 J
*824. A 12lb hammer moving at 80 ft/s strikes the head of a nail moving it into the wall a distance of ¼ in. What was the average stopping force?
s = 0.250 in. (1 ft/12 in.) = 0.0203 ft; v_{o} = 80 ft/s
Fs = ½mv_{f}^{2}  ½mv_{o}^{2}; F (0.0203 ft) = 0  ½(0.375 slugs)(80 ft/s)^{2}; F = 57,600 ft lb
825. What average force is needed to increase the velocity of a 2kg object from 5 m/s to 12 m/s over a distance of 8 m?
Fs = ½mv_{f}^{2}  ½mv_{o}^{2}; F(8 m) = ½(2 kg)(12 m/s)^{2}  ½(2 kg)(5 m/s)^{2}; F = 14.9 N
*826. Verify the answer to Problem 825 by applying Newton’s second law of motion.
To apply F = ma, we need to find a: 2as = v_{f}^{2} – v_{o}^{2}
F = (2 kg)(7.44 m/s^{2}) = 14.9 N
*827. A 20g projectile strikes a mud bank in Fig. 812, penetrating a distance of 6 cm before stopping. Find the stopping force F if the entrance velocity is 80 m/s.
Fs = ½mv_{f}^{2}  ½mv_{o}^{2}; F (0.06 m) = 0  ½(0..02 kg)(80 m/s)^{2}
F = 1070 N
*828. A 1500kg car is moving along a level road at 60 km/h. What work is required to stop the car? If _{k} = 0.7, what is the stopping distance? (60 km/h = 16.67 m/s)
Work = ½mv_{f}^{2}  ½mv_{o}^{2} ; Work = 0  ½(1500 kg)(16.67 m/s)^{2}; Work = 208,300 J
The work is done by friction: F = _{k}N = _{k} mg and (Work)_{F}= (_{k} mg)s
(_{k} mg)s = 208,300 J; ; s = 20.2 m
Potential Energy
829. A 2kg block rests on top of a table 80 cm from the floor. Find the potential energy of the book relative to (a) the floor, (b) the seat of a chair 40 cm from the floor, and (c) relative to the ceiling 3 m from the floor?
For E_{p }= mgh, the height h is measured from reference point:
For floor, h = 0.8 m; for seat, h = 0.4 m; for table, h =  2.2 m
(a) E_{p} = (2 kg)(9.8 m/s^{2})(0.8 m) = 15.7 J
(b) E_{p} = (2 kg)(9.8 m/s^{2})(0.4 m) = 7.84 J ; (c) E_{p} = (2 kg)(9.8 m/s^{2})(2.2 m) = 43.1 J
830. A 1.2 kg brick is held a distance of 2 m above a manhole. The bottom of the manhole is 3 m below the street. Relative to the street, what is the potential energy at each location? What is the change in potential energy?
E_{p} = (1.2 kg)(9.8 m/s^{2})(2 m) = 23.5 J for held brick.
E_{p} = (1.2 kg)(9.8 m/s^{2})(3 m) = 35.3 J for brick in hole
E_{p} = E_{f} – E_{o} = 35.3 J – (23.5 J); E_{p} = 58.8 J
831. At a particular instant a mortar shell has a velocity of 60 m/s. If its potential energy at that point is onehalf of its kinetic energy, what is its height above the earth?
E_{k} = ½mv^{2} and E_{p} = mgh; At the instant in question, E_{p} = ½E_{k}
and
h = 91.8 m
*832. A 20kg sled is pushed up a 34^{0} slope to a vertical height of 140 m. A constant friction force of 50 N acts for the entire distance. What external work was required? What was the change in potential energy?
External work done by force P acting for distance s:
F_{x} = 0: P – mg sin 34^{0} – F = 0; F = 50 N, m = 20 kg
P = (5 kg)(9.8 m/s^{2}) sin 34^{0} + 50 N; P = 77.4 N
; (Work)_{P} = (77.4 N)(250 m); (Work)_{P} = 19.400 J
E_{p} = mgh = (5 kg)(9.8 m/s^{2})(140 m); E_{p} = 6860 J
The difference: 19,400 J – 6860 J = 12,540 J is the work done against friction.
*
0.04 m
600 N
833. An average force of 600 N is required to compress a coiled spring a distance of 4 cm. What work is done BY the spring? What is the change in potential energy of the compressed spring?
Work done BY spring is opposite to compressing force.
Work = (600 N)(0.04 m) = 24.0 J Work by spring =  24.0 J
Now, E_{P} = 0 initially, so that E_{p} = (Work)_{SP} = (24 J); E_{p} = +24.0 J
Conservation of Energy (No Friction)
834. A 64lb weight is lifted to a height of 10 ft and then released to fall freely. What is the potential energy, the kinetic energy and the total energy at (a) the highest point, (b) 3 ft above the ground, and (c) at the ground? (W = 64 lb, g = 32 ft/s^{}
(a) E_{P} = Wh = (64 lb)(10 ft) = 640 ft lb; E_{P} = 640 ft lb
E_{k} = ½mv^{2} = 0 ( v_{o} = 0); E_{T} = E_{P} + E_{k} = 640 ft lb + 0
At 10 ft: E_{P}_{ }= 640 ft lb; E_{k} = 0; and E_{T} = 640 ft lb
(b) E_{P} = Wh = (64 lb)(3 ft) = 192 ft lb; E_{P} = 192 ft lb
E_{k} = E_{T} – E_{P}; E_{k} = 640 ft lb – 192 ft lb; E_{k} = 448 ft lb
At 3 ft: E_{P}_{ }= 192 ft lb; E_{k} = 448 ft lb; and E_{T} = 640 ft lb
(c) At 0 ft h = 0 and E_{T} is same: E_{P}_{ }= 0 ft lb; E_{k} = 640 ft lb; and E_{T} = 640 ft lb
835. A 4kg hammer is lifted to a height of 10 m and dropped? What are the potential and kinetic energies of the hammer when it has fallen to a point 4 m from the earth?
At 10 m: E_{k} = 0 and E_{P} = mgh = (4 kg)(9.8 m/s^{2})(10 m); E_{T} = 0 + 392 J = 392 J
E_{p} = mgh = (4 kg)(9.8 m/s^{2})(4 m); E_{p }= 157 J; E_{k} = E_{T} – E_{P} = 392 J – 157 J
Thus at h = 4 m: E_{P} = 157 J and E_{k} = 235 J
*836. What will be the velocity of the hammer in Problem 835 just before striking the ground? What is the velocity at the 4m location?
At bottom, E_{p} = 0 and E_{T} = 392 J so that E_{k} = 392 J – 0 = 392 J;
E_{k} = ½mv^{2} = 392 J v = 14.0 m/s
*837. What initial velocity must be given to a 5kg mass if it is to rise to a height of 10 m? What is the total energy at any point in its path?
In absence of friction, total energy at bottom must equal total energy at top:
E_{T} = ½mv_{o}^{2} + 0 = 0 + mgh; v_{o}^{2} = 2 gh
v_{0} = 14.0 m/s^{2} E_{T} = E_{p} + E_{k} at any point
At top: E_{T} = 0 + mgh = (5 kg)(9.8 m/s^{2})(10 m); E_{T } = 490 J
*838. A simple pendulum 1 m long has an 8kg bob. How much work is needed to move the pendulum from its lowest point to a horizontal position? From energy considerations find the velocity of the bob as it swings through the lowest point.
A force F equal to the weight mg must act through a distance equal to length of string:
Work = mgh = (8 kg)(9.8 m/s^{2})(1 m); Work = 78.4 J
The total energy at top(mgh) must be equal to total energy at bottom.(½mv^{2}):
mgh = ½mv^{2 } v = 4.43 m/s
*839. A ballistic pendulum is illustrated in Fig. 813. A 40g ball is caught by a 500g suspended mass. After impact, the two masses rise a vertical distance of 45 mm. Find the velocity of the combined masses just after impact? (See Figure next page)
Total mass M = 40 g + 500 g = 540 g; M = 540 g = 0.540 kg
*839. (Cont.) Find v_{o} of total mass M such that M rises h = 0.045 m:
Energy conservation: ½Mv^{2} + 0 = 0 + Mgh;
v = 0.939 m/s
*
h
840. A 100lb sled slides from rest at the top of a 37^{0} inclined plane. The original height is 80 ft. In the absence of friction, what is the velocity of the sled when it reaches the bottom of the incline? (Not dependent on either angle or weight.)
Energy conservation: 0 + mgh = ½mv^{2} + 0;
v = 71.6 ft/s
*841. An 8kg block in Fig. 814 has an initial downward velocity of 7 m/s. Neglecting friction, find the velocity when it reaches point B?
½mv_{o}^{2} + mgh_{o} = ½mv_{f}^{2} + mgh_{f}
v_{o}^{2} + 2gh_{o} = v_{f}^{2} + 0 ; v_{f}^{2} = ^{ }v_{o}^{2} + 2gh_{o}
; v_{f} = 21.0 m/s
*842. What is the velocity of the 8kg block at point C in Problem 839? (Note h_{f} 0 this time)
½mv_{o}^{2} + mgh_{o} = ½mv_{f}^{2} + mgh_{f };_{ } v_{o}^{2} + 2gh_{o} = v_{f}^{2} + 2gh_{f} ; v_{f}^{2} = ^{ }v_{o}^{2} + 2gh_{o} – 2gh_{f}
; v_{f} = 16.9 m/s
*843. An 80lb girl sits in a swing of negligible weight. If she is given an initial velocity of 20 ft/s, to what height will she rise?
Energy conservation: 0 + mgh = ½mv^{2} + 0;
h = 6.25 ft
Energy and Friction Forces
*844. A 60kg sled slides to the bottom of a 25^{0} slope of length 30 m. A 100N friction force acts for the entire distance. What is the total energy at the top of the slope and at the bottom? What is the velocity of the sled at the bottom?
h = (20 m)sin 25^{}m; At top: E_{P} = mgh; E_{k} = 0
E_{T} = E_{P} + E_{k} = mgh + 0; E_{T} = (60 kg)(9.8 m/s^{2})(8.45 m)
Total energy at top, E_{T} = 4969 J Cons. of E: E_{T}(top) = E_{kf} (bottom)+Losses
Loss = (Work)_{F} = F s; Loss = (100 N)(30 m) = 3000 J
Cons. of E: 4960 J = ½(60 kg)v^{2} + 3000 J; From which: v = 8.10 m/s
*845. A 500g block is released from the top of a 30^{0} inline and slides 160 cm to the bottom. A constant friction force of 0.9 N acts the entire distance. What is the total energy at the top? What work is done by friction? What is the velocity at the bottom?
W = mg = (0.5 kg)(9.8 m/s^{2}) = 4.90 N;
h = (1.60 m) sin 30^{0} = 0.800 m; E_{T} = E_{P} + E_{k}
E_{T }= Wh + 0; E_{T} = (4.90 N)(0.80 m); E_{T} = 3.92 J
(Work)_{F} = F s = (0.900 N)(1.60 m); (Work)_{F} = 1.44 J (negative work)
Total energy at top = total energy at bottom + work done against friction
3.92 J = ½mv^{2} + F s ; 3.92 J = ½(0.5 kg) v^{2} + 1.44 J
Solving for v, we obtain: v = 3.15 m/s
Note that the work done BY friction is negative, but when applying conservation of energy we use the work AGAINST friction (+1.44 J) to account for the LOSS
*846. What initial velocity must be given to the 500g block in Problem 843 if it is to just reach the top of the same slope? (See previous problem)
From Prob. 843: F = 0.9 N, h =0.8 m, W = 4.90 N
½mv_{o}^{2} = Wh_{f} + F s = 3.92 J + 1.44 J ; ½mv_{o}^{2} =5.36 J
v_{o} = 4.63 m/s
*847. A 64lb cart starts up a 37^{0} incline with an initial velocity of 60 ft/s. If it comes to rest after a moving a distance of 70 ft, how much energy was lost to friction?
W = 64 lb; m = (64/32) = 2 slugs; h = 70 sin 37^{}m
½mv_{o}^{2} = Wh_{f} + Loss; Loss = ½mv_{o}^{2}  Wh_{f}
Loss = ½(2 slugs)(60 ft/s)^{2} – (64 lb)(42.1 ft)
Loss = 3600 ft lb  2240 ft lb; Loss = 904 ft lb
*848. A 0.4kg ball drops a vertical distance of 40 m and rebounds to a height of 16 m. How much energy was lost in collision with the floor? Conservation of energy.
mgh_{o} = mgh_{f} + Loss; Loss = mgh_{o} – mgh_{f} = mg(h_{o} – h_{f})
Loss = (0.4 kg)(9.8 m/s^{2})(40 m – 16 m); Loss = 94.1 J
*849. A 4kg sled is given an initial velocity of 10 m/s at the top of a 34^{0} slope. If _{k} = 0.2, how far must the sled travel until its velocity reaches 30 m/s?
mgh_{o} + ½mv_{o}^{2} = 0 + ½mv_{f}^{2} + F s and h_{o} = s sin 34^{0}
mg(s sin 34^{0})  F s = ½mv_{f}^{2 } ½mv_{o}^{2} ; F = _{k} mg cos 34^{0 }
mg(s sin 34^{0}) – (_{k}mg cos 34^{0})s^{ }= ½mv_{f}^{2 } ½mv_{o}^{2}
*849. (Cont.) (g sin 34^{0}  _{k}g cos 34^{0})s = ½v_{f}^{2 } ½v_{o} ;
s = 104 m
*850. Assume in Fig. 814 that the sliding mass is 6kg and that 300 J of energy is lost doing work against friction. What is the velocity when the mass reaches point C?
½mv_{o}^{2} + mgh_{o} = ½mv_{c}^{2} + mgh_{f} + 300 J
½mv_{c}^{2} = ½mv_{o}^{2} + mgh_{o} – mgh_{c} – 300 J
v_{c}^{2} = v_{o}^{2} + 2g(h_{o} – h_{c}) 
v_{c} = 13.6 m/s
*851. A bus slams on brakes to avoid an accident. The tread marks of the tires are 80 feet long. If _{k} = 0.7, what was the speed before applying brakes?
0
F = _{k}mg; Work = F s = _{k}mgs
Work = E_{k} ; _{k}mgs = ½mv_{f}^{2}  ½mv_{o}^{2}
v_{o} = 59.9 ft/s
Power
852. A powerstation conveyor belt lifts 500 tons of ore to a height of 90 ft in one hour. What average horsepower is required? (W = 500 tons = 1 x 10^{6} lb; 1 hp = 550 ft lb/s)
; P = 25,000 ft lb/s = 45.5 hp
853. A 40kg mass is lifted through a distance of 20 m in a time of 3 s. Find average power.
; P = 2610 W
854. A 300kg elevator is lifted vertical distance of 100 m in 2 min. What is the output power?
; P = 2.45 kW
855. A 90 kW engine is used to lift a 1200kg load. What is the average velocity of the lift?
; ; v _{ }= 7.65 m/s
856. To what height can a 400 W engine lift a 100kg mass in 3 s?
; h = 0.122 m
857. An 800N student runs up a flight of stairs rising 6 m in 8 s. What is the average power?
; P = 600 W
*858. A speedboat must develop a 120 hp in order to move at a constant speed of 15 ft/s through the water. What is the average resistive force due to the water?
; F = 4400 lb
Challenge Problems
*859. A worker lifts a 20kg bucket from a well at constant speed and does 8 kJ of work. How deep is the well?
Work = Fs ; ; s = 40.8 m
*860. A horizontal force of 200 N pushes an 800N crate horizontally for a distance of 6 m at constant speed. What work is done by the 200N force. What is the resultant work?
Work = Fs = (200 N)(6 m); Work = 1200 J
Speed is constant, so F_{R} = 0, and Resultant work = 0 J
*861. A 10kg mass is lifted to a height of 20 m, and released. What is the total energy of the system? What is the velocity of the mass when it is located 5 m from the floor?
In absence of friction, total energy is constant, so that: E_{T} (Top) = E_{T}(5 m)
E
_{T}_{ } = mgh + 0 = (10 kg)(9.8 m/s^{2})(20 m); E_{T} = 1960 J
When h = 5 m, (10 kg)(9.8 m/s^{2})(5 m) + ½(10 kg)v_{f}^{2} = 1960 J
490 J + (5 kg)v_{f}^{2} = 1960 J; v_{f} = 17.1 m/s
*862. A crate is lifted at a constant speed of 5 m/s by an engine whose output power is 4 kW. What is the mass of the crate?
; m = 81.6 kg
*863. A roller coaster boasts a maximum height of 100 ft. What is the maximum speed in miles per hour when it reaches its lowest point? (Conservation of Energy)
mgh + 0 = 0 + ½mv^{2} ; ; v = 80.0 ft/s or 54.4 mi/h
*864. A 20N force drags an 8kg block a horizontal distance of 40 m by a rope at an angle of 37^{0} with the horizontal. Assume _{k} = 0.2 and that the time required is one minute. What resultant work is done?
Resultant work = work of the resultant force
F_{R} = (20 N)cos 37^{}  F) F = _{k}N F_{y} = 0
N + 20 sin37^{0} – (8 kg)9.8 m/s^{2}) = 0; N = 66.4 N; F = 0.2(66.4 N) = 13.3 N
F_{R} = 20 cos 37^{0} – 13.3 N = 2.70 N; Work = F_{R}s = (2.70 N)(40 m); Work = 108 J
*865. What is the velocity of the block in Problem 860 at the end of the trip? What resultant power was expended? (Assume block starts from rest, then apply workenergy theorem.)
Work = ½mv_{f}^{2}  ½mv_{o}^{2} ; 108 J = ½(8 kg)v^{2} + 0; ; v = 5.20 m/s
P = 1.80 W
*866. A 70kg skier slides down a 30 m slope that makes an angle of 28^{0} with the horizontal. Assume that _{k} = 0.2. What is the velocity of the skier at the bottom of the slope?
0
mgh_{o} + ½mv_{o}^{2} = 0 + ½mv_{f}^{2} + F s and h_{o} = s sin 28^{0}
mg(s sin 28^{0})  F s = ½mv_{f}^{2 } ½mv_{o}^{2} ; F = _{k} mg cos 28^{0 }
mg(s sin 28^{0}) – (_{k}mg cos 28^{0})s^{ }= ½mv_{f}^{2 }
*866. (Cont.) (g sin 28^{0}  _{k}g cos 28^{0})s = ½v_{f}^{2};
v_{f} = 13.1 m/s
*867. A 0.3 mg flea can jump to a height of about 3 cm. What must be the takeoff speed? Do you really need to know the mass of the flea?
½
mv_{o}^{2} = mgh ; ;
v = 0.767 m/s ; The mass is not needed.
*868. A roller coaster goes through a low point and barely makes the next hill 15 m higher. What is the minimum speed at the bottom of the loop?
½
mv_{o}^{2} = mgh ; ; v = 17.1 m/s
*869. The hammer of a pile driver weighs 800 lb and falls a distance of 16 ft before striking the pile. The impact drives the pile 6 in. deeper into the ground. What was the average force driving the pile? m = W/g = (800/32) = 25.0 slugs; s = 6 in. = 0.5 ft
The work done by the pile driver Fs is determine from the change in kinetic energy, so we need to find the velocity of the driver just before striking the stake:
½mv_{f}^{2} = mgh_{o} ; ; v = 32.0 ft/s
Work to stop driver = change in kinetic energy of driver
Fs = ½mv_{f}^{2}  ½mv_{o}^{2}; F = 25,600 lb
Critical Thinking Questions
*870. An inclined board is used to unload boxes of nails from the back of a truck. The height of the truck bed is 60 cm and the board is 1.2 m in length. Assume that _{k }= 0.4 and the boxes are given an initial push to start sliding. What is their speed when they reach the ground below. What initial speed would they need at the bottom in order to slide back into the truck bed? In the absence of friction would these two questions have the same answer? [ h = 0.6 m; s = 1.2 m ; sin = 0.6/1.2; = 30^{0} ]
(a) (Work)_{F} = F s = _{k}N = _{k} mg cos 30^{0} s
mgh = ½mv^{2} + _{k} mg cos 30^{0} s; 2gh = v^{2} +2_{k}gs cos 30^{0}
v^{2} = 2gh – 2g_{k} s cos 30^{0} = 2(9.8 m/s^{2})(0.6 m) – 2(9.8 m/s^{2})(0.4)(1.2 m)(0.866)
v^{2} = 11.76 – 8.15 = 3.61 m^{2}/s^{2}; ; v = 1.90 m/s
(b) Going up the plane, the initial speed must provide the energy to overcome the
friction force which would now be directed DOWN the plane.
½mv^{2} = mgh + _{k}mg cos 30^{0} s (Note the difference in this equation from that above.)
v^{2 }= 2gh +2_{k}gs cos 30^{0} = (2)(9.8)(0.6) + 2(0.4)(9.8)(1.2)(0.866)
v^{2} = 11.76 + 8.15 = 19.9 m^{2}/s^{2}; ; v = 4.46 m/s
In the downhill case, the initial potential energy was lost to friction and what little remained appeared in the form of a small velocity at the bottom. In the uphill case, the initial kinetic energy (high velocity) was used to gain the height h, but more energy was needed to overcome friction. In the absence of friction, height is transferred into velocity going down, and velocity is transferred to height going up. Thus, absent friction, the same velocities would be found for each of the above cases. (v = 3.43 m/s)
*
871. A 96lb safe is pushed with negligible friction up a 30^{0} incline for a distance of 12 ft. What is the increase in potential energy? Would the same change in potential energy occur if a 10lb friction force opposed the motion up the incline? Why? Would the same work be required? [ h = 12 sin 30^{0} = 6.00 ft ]
(a) E_{p} = Wh = (96 lb)(6 ft) E_{p} = 576 ft lb
(b) E_{p} is a function only of weight and height, so the same change
in potential energy occurs regard less of friction or the path taken.
(c) With a 10lb_{ }friction force, a work of (10 lb)(12 ft) = 120 ft lb is needed in addition
to the work of 576 ft lb just to lift the weight. The total work is 696 ft lb.
*872. A 2kg ball is suspended from a 3m cable attached to a spike in the wall. The ball is pulled out, so that the cable makes an angle of 70^{0} with the wall, and then released. If 10 J of energy are lost during the collision with the wall, what is the maximum angle between the cable and the wall after the first rebound?
y_{o} = (3 m) cos 70^{0} = 1.026 m
h_{o} = 3 m – 1.026 m; h_{o} = 1.974 m
mgh_{o} = mgh_{f }+ 10 J; mgh_{f} = mgh_{o} – 10 J
; h_{f} = 1.464 m ; y_{f} = 3 m – 1.464 m = 1.536 m
; = 59.2^{0}
*873. A 3kg ball dropped from a height of 12 m has a velocity of 10 m/s just before hitting the ground. What is the average retarding force due to the air? If the ball rebounds from the surface with a speed of 8 m/s, what energy was lost on impact? How high will it rebound if the average air resistance is the same as before?
First apply E conservation to the falling portion of the problem:
mgh_{o} = ½mv^{2} + F s ; F s = mgh_{o}  ½mv^{2}
F (12 m) = (3 kg)(9.8 m/s^{2})(12 m)  ½(3 kg)(10 m/s)^{2}; F = 16.9 N
The loss or work done on impact equals the change in E_{k}: Loss = Work = ½mv_{f}^{2}  ½mv_{o}^{2}
Work = ½(3 kg)(8 m/s)^{2}  ½(3 kg)(10 m/s)^{2}; Work =  54 J; Impact loss = 54 J
To find rebound height, apply conservation of energy with losses to air and to impact:
mgh_{o} = mgh_{f} + (F s)_{Air} + Impact loss; ( s = 12 m + h_{f} )
(3 kg)(9.8 m/s^{2})(12 m) = (3kg)(9.8 m/s^{2})h_{f}_{ }+ (16.9 N)(12 m + h_{f}) + 54 J
353 J = (29.4 N)h_{f} + 203 J + (16.9 N)h_{f} + 54 J; h_{f} = 2.07 m
*
874. Consider a roller coaster where the first hill is 34 m high? If the coaster losses only 8% of its energy between the first two hills, what is the maximum height possible for the second hill?
mgh_{o} = mgh_{f} + 0.08 mgh_{o}; h_{f} = (1 – 0.08)h_{o }= 0.92 (34 m) h_{f} = 31.3 m
*875. A 4kg block is compressed against a spring at the bottom the inclined plane in Fig. 815. A force of 4000 N was required to compress the spring a distance of 6 cm. If it is then released and the coefficient of friction is 0.4, how far up the incline will the block move?
Work to compress spring = (4000 N)(0.06 m) = 240 J = E_{p}
E_{p}(spring) = mgh_{ }+ F s; F s = _{k}mg cos 30^{0} s
F s = (0.4)(4 kg)(9.8 m/s^{2})(0.866) s = (13.6 N) s
240 J = (4 kg)(9.8 m/s^{2})h + (13.6 N) s h = s sin 30^{0} = 0.5 s
240 J = (4 kg)(9.8 m/s^{2})(2 s) + (13.6 N) s; s = 2.61 m
