Answer:
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 work in J done by the force of gravity on the ball while the ball is being thrown is -4.95.
Since 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.
Now the work done is
= -mgs
= -0.5 * 9.8 * 1m
= -4.9 J
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