Question

A 5 kg body is fired vertically up with a speed of-.200 m/s. Just before it hits the ground, its speedis 150 m/s. Over the entire trip, the work done bygravity is(1) 45000 J(2) – 43750 J(3) 42000(4) Zero​

Answer 1

Answer:

it is 0 because gravity is a conservative force and according to the condition if you try to find the net displacement then it is clear that a body returns to its initial position and so the net displacement is zero so work done by gravity will be zero but some amount of work will be done by air resistance that's why the body does not have it initial velocity on returning back to the observer

Answer 2

Answer:

Net work done by gravity is zero.

Given:

A 5 kg body is fired vertically up with a speed of-.200 m/s. Just before it hits the ground, its speed is 150 m/s.

To Find:

Find the work done by gravity over the trip.

Explanation:

For the upward trip:

Force acting on the body by gravity = -mg N

=5x10 = 50N

-ve sign shows that force is acting opposite to motion.

initial velocity ,  u = 200m/sec

final velocity , v = 0 m/sec

v² = u² + 2aS

0= u² -2gS

2 x 10 x S = 200²

S = 40000/20

S = 2000 m

Work done by gravity =- FxS

=-50x2000

=-100000 J

Downward trip:

Force applied = 50 N

and the same distance will be covered = 2000m

So work done by gravity = 50x2000

=100000 J

So net work done by gravity = 100000-100000

=0 Joules

So net work done by gravity during the whole trip is zero, but after returning to the ground the speed is nor 200 m/sec because some work is done by air resistance , which reduces the speed of the body.