Question

A 200. N wagon is to be pulled up a 30 degree incline at constant speed. How large a force parallel to the incline force is needed? Assume no friction. ...?

Answer 1

For the question above, here is the equation to follow:
F = mgsinα = Wsinα
=200 x 0.5 = 100 N
OR

Sin30 * 200N = 100 N

The asnwer is 100N. I hope this answer helps.

Answer 2

F = 100 N

Step-by-step explanation:

Since the wagon is pulled along the inclined plane upwards so here the component of the weight of the wagon will act down the plane.

There is no friction force on the wagon so here in order to move the wagon upwards along the plane we require a force along the plane upwards which must be of same magnitude as that the magnitude of weight along the plane downwards.

now the component of weight along the inclined is given as

`F = Wsin\theta`

`F = 200 sin30`

`F = 200(0.5)`

`F = 100 N`