ANSWERS
(a) 1000 J
(b) 2000 J
(c) 50%
EXPLANATION
Given:
• The weight of the box, Fg = 500 N
,
• The length of the plank, l = 10 m
,
• The vertical height of the truck, h = 2 m
,
• The force applied to pull the box, F = 200 N
Find:
• (a), The change in gravitational potential energy, PE
,
• (b), The work done by the 200 N force, W
,
• (c), The efficiency of the set up
(a) The change in gravitational potential energy is,
[tex]PE=mgh[/tex]
The product mg is the weight of the box,
[tex]PE=500N\cdot2m=1000J[/tex]
Hence, the change in gravitational potential energy of the box is 1000 J.
(b) The 200 N force is applied along the plank, so the distance it is applied for is the length of the plank and, therefore, the work done by this force is,
[tex]W=200N\cdot10m=2000J[/tex]
Hence, the work done by the 200 N force is 2000 J.
(c) The efficiency is the quotient between the output work - which is the work needed to do the job, and the input work - which is the work done,
[tex]efficiency=\frac{W_{output}}{W_{input}}\cdot100=\frac{1000J}{2000J}\cdot100=0.5\cdot100=50\%[/tex]
Hence, the efficiency is 50%.