11. Ws = 5000 N. = Normal force(Fn).
Fk = u*Fn = 0.05 * 5000 = 250 N. = Eorce
of kinetic friction.
Fap-Fk = M*a
Fap - 250 = M*0 = 0
Fap = 250 N. = Force applied.
Work = Fap * d = 250 * 1000 = 250,000 J.
= 2.5*10^5 J.
12. Same procedure as #11.13.
13. Ws = 6000 N. = Normal force(Fn).
Fk = u*Fn = 0.05 * 6000 = 300 N.
M*g = 6000
M * 9.8 = 6000
M = 612 kg
Fap-Fk = M*a
Fap - 300 = 612*0.20
Solve for Fap.
Work = Fap * d
14. Wb = F = M*g = 1 * 10 = 10 N.
Work = F*d
10. A 2-kg object is moving at 3m/s. A 4-N force is applied in the direction of motion and then
removed after the object has traveled an additional 5m. The work done by this force is:
A. 12 J
B. 15 J
C. 18 J
D. 20 J
E. 38 J
ans: D
11. A sledge (including load) weighs 5000 N. It is pulled on level snow by a dog team exerting a
horizontal force on it. The coefficient of kinetic friction between sledge and snow is 0.05. How
much work is done by the dog team pulling the sledge 1000m at constant speed?
A. 2.5 × 104 J
B. 2.5 × 105 J
C. 5.0 × 105 J
D. 2.5 × 106 J
E. 5.0 × 106 J
ans: B
12. Camping equipment weighing 6000N is pulled across a frozen lake by means of a horizontal
rope. The coefficient of kinetic friction is 0.05. The work done by the campers in pulling the
equipment 1000m at constant velocity is:
A. 3.1 × 104 J
B. 1.5 × 105 J
C. 3.0 × 105 J
D. 2.9 × 106 J
E. 6.0 × 106 J
ans: C
13. Camping equipment weighing 6000N is pulled across a frozen lake by means of a horizontal
rope. The coefficient of kinetic friction is 0.05. How much work is done by the campers in
pulling the equipment 1000m if its speed is increasing at the constant rate of 0.20m/s2?
A. −1.2 × 106 J
B. 1.8 × 105 J
C. 3.0 × 105 J
D. 4.2 × 105 J
E. 1.2 × 106 J
ans: D
14. A 1-kg block is lifted vertically 1m by a boy. The work done by the boy is about:
A. 1 ft · lb
B. 1 J
C. 10 J
D. 0.1J
E. zero
ans: C
88 Chapter 7: KINETIC ENERGY AND WORK
1 answer