Gravitational Potential energy of a box and a slope

Question

A 20 kg box slides 45 m from the top down to the bottom of an incline which is at an angle of 10° to the horizontal. The box is stationary at the top of the slope and is accelerating down the slope at a rate of 1.0 m s^-2

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a) How much work is done by the force of gravity on the box?

b) How much work is done by the frictional force?

c) What is the kinetic energy of the box at the bottom of the slope?

d) How much gravitational potential energy has the box lost while traveling down the slope?

Answer 1

A 20 kg box slides 45 m from the top down to the bottom of an incline which is at an angle of 10° to the horizontal. The box is stationary at the top of the slope and is accelerating down the slope at a rate of 1.0 m s^-2

m=20 kg

D=45 m

a=1.0 m/sec²

Θ=10^o

a) How much work is done by the force of gravity on the box?

None. The only energy terms are kinetic energy (KE) due to the motion of the box, potential energy (PE) due to the height of the box, and work (W) done by friction.

b) How much work is done by the frictional force?

The elevation of the box is DsinΘ so the potential energy

PE =mgDsinΘ

PE=(20)(9.81)(45)sin(10^o)= 1533 J

Along the slope, s=s(0)+v(0)t+0.5at²

v(0)=0

s(0)=0

a=1 m/sec²

s=D at time t

45=0.5(1 m/sec²)t²

t=√90=9.487 sec

v=v(0)+at=(1)9.487=9.487 m/sec

KE=0.5mv²=0.5(20 kg)9.487²=900 J

Work along slope=W=PE-KE=1533-900=633 J

Solution: 0.6 kJ

c) What is the kinetic energy of the box at the bottom of the slope?

Solution: 0.9 kJ

d) How much gravitational potential energy has the box lost while traveling down the slope?

Solution: 1.5 kJ