A mass exerts 45.0N on a 30 kg box, making it slide over a smooth surface for a distance of 10.0m . He stops pushing, just as the box crosses the 10.0m mark.
(a) what is the work done by the man on the box?
(b) what form of energy does the box posses after the end of the 10.0m, when the man has stopped pushing ?
(c) what is the velocity of the Vic after the man stops pushing ?
(d) how would your answer to (c) change of the same box was pushed (with the same force) over a rough surface instead ?
(e) give reasons to support your answer in (d)
(a) what is the work done by the man on the box?
Force * distance = 45 N * 10 meters = 450 Joules
(b) what form of energy does the box posses after the end of the 10.0m, when the man has stopped pushing ?
Kinetic energy = (1/2) m v^2
(c) what is the velocity of the Vic after the man stops pushing ?
well, I just told you (1/2) mv^2 = energy gained = work done = 450 Joules
so
(1/2) (30) v^2 = 450
v^2 = 900/30 = 30*30/30 = 30
v = sqrt(30) meters/second
(d) how would your answer to (c) change of the same box was pushed (with the same force) over a rough surface instead ?
(e) give reasons to support your answer in (d)
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ah come on. If there is friction force opposing your 45 Newtons, the net force accelerating the box is less so it will not accelerate as much. The lost work you do is lost as heat.