Question

A 3-kg object is moving at 5.0 m/s. An 12-N force is applied in the direction of motion and then removed after the object has traveled an additional 6.0 m. What is the work done by this force?

Answer 1

The work done by the force is 109.5 Joules.

Step-by-step explanation:

It is given that,

Mass of the object, m = 3 kg

Speed of the object, v = 5 m/s

Force applied on the object, F = 12 N

Distance covered by the object, d = 6 m

The work done by this force is given by the sum of kinetic energy and potential energy as per work energy theorem as :

`W=K+P`

`W= (1)/(2)mv^2+Fd`

`W=(1)/(2)* 3* (5)^2+12* 6`

W = 109.5 Joules

So, the work done by the force is 109.5 Joules. Hence, this is the required solution.