A person throws a 0.5 kg ball up in the air in 1.1 s. During this process the position of the ball changes from 0.5 m to 1.5 m (before being released) and the speed of the ball changes from 0 m/s to 7 m/s. What is the work (in J) done by the force of gravity on the ball while the ball is being thrown?

Explanation:

A ball is thrown up and it works against gravity

Mass = 0.5kg

Time 1.1sec

Change of position ∆h=0.5 to 1.5m

∆h=1.5-0.5

∆h=1m

Work done by gravity is given as=W×d

W=mg, let g, =9.8m/s²

W=0.5×9.8

W=4.9N

Then the upward motion

Initial velocity 7m/s

Final velocity 0m/s

So this final velocity will be the initial velocity of the upward motion

We need to get the height, the ball traveled

S=ut-1/2gt² motion against gravity

u=7m/s

S=7×1.1-0.5×9.8×1.1²

S=1.771m

The height the ball reached is 1.771m

The total height of the ball from start is ∆h +S

H=1.771+1

H=2.771m

Work done done by gravity =W×H

Workdone= 4.9 × 2.771

Work done=13.58Joules

But the work done from when the ball is release up to reach the maximum height is given as

Work done = W×s

Work done =4.9× 1.771

Work done = 8.68J

This is the work done by gravity during the thrown alone, when the ball left the hand

the answer is b

explanation: