Question

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. Over the entire trip, the work done by
gravity is
(1) 45000 J
(2) – 43750 J
(3) 42000
(4) Zero​

Answer 1

Answer:

W = -43750 J

Explanation:

Given that,

Mass of the body, m = 5 kg

Initial speed of the body, u = 200 m/s

Final speed of the body, v = 150 m/s

Let W is the work done by the gravity. It is equal to the change in kinetic energy. Its formula is given by :

$W=\dfrac{1}{2}m(v^2-u^2)$

$W=\dfrac{1}{2}\times 5\times ((150)^2-(200)^2)$

W = -43750 J

So, the work done by  gravity over the entire trip is -43750 J. Hence, this is the required solution.