Question

8. An object (mass = 200 kg) is pushed up a friction- less inclined plane (0) by a force (F) that is parallel to the incline. At the start of an 8 m displacement (d), the object's speed is 15 m/sec and at the end it is 25 m/sec. We are told that the work done by gravity on the ob- ject during the 8 m displacement is (-5000 J); what must be the magnitude of the force (F)? (in N) a. 6250 5625 c. 7500

Answer 1

By work energy theorem

Wnet = UK

= (1/2)m(v² - u²)

Wf+W,

(F×8)+(-5000) = (1/2)×200×(25² -15²)

8F -5000 = 40000

8F = 45000

F = 5625 N

(Note that work done by applied force is equal to force multiplied by displacement as both are in same direction.

$\Delta K$ is change in kinetic energy)