Question

A 60.0 kg object is moving east at 8.00 m/s, and then slows down to 4.00 m/s. How much work was done?

Answer 1

The answer is  -1440. That is letter A.

Answer 2

Answer:

Work done, W = -1440 J

Explanation:

It is given that,

Mas of an object, m = 60 kg

Initial velocity of the object, u = 8 m/s

Final velocity of an object, v = 4 m/s

We know that work done is equal to change in kinetic energy of an object as per work energy theorem.

$W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

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

Putting all the values in above equation we get :

$W=\dfrac{1}{2}\times 60\ kg((4\ m/s)^2-(8\ m/s)^2)$

W = -1440 J

So, the work done by the object is -1440 Joules. Hence, the correct option is (A) " -1440 J ".