Question

A3kg crate is placed at the bottom of an inclined frictionless plane making an angle 30 ° above the horizontal and pushed up by a force F - 50N parallel to the incline as shown below . If the incline is 6 m long , then the net work done on this crate when it reaches the top of the incline is equal to :

Answer 1

Answer: 210 J

Step-by-step explanation:

Since there is no friction, coefficient of friction, μ = 0

The free body diagram is shown below

There must be a downward force due to gravity. Thus,

Net force = applied force - downward force

From the diagram,

Downward force = mgSinθ

where

m is the mass of the crate

θ is the angle between the plane and the horizontal axis

g is the acceleration due to gravity

From the information given,

m = 3

applied force = 50N

θ = 30

g = 10 m/s^2

Net force = (50 - 3 x 9.81Sin30) = 35 N

Net work = net force x distance

Given that distance = 6m,

Net work = 35 x 6

Net work = 210 J