# How much work does it take to push an object with a mass of 2 kg up a 7 m ramp, if the ramp has an incline of (11pi)/12  and a kinetic friction coefficient of 8 ?

Feb 4, 2018

Inclination of $\frac{11 \pi}{12} = 165 \mathrm{de} g r e e s$ means with the other direction of the horizontal level,the inclined plane makes an angle of $\left(180 - 165\right) = 15 \mathrm{de} g r e e s$

So,the component of weight of the object which tries to pull it down along the inclined plane is $m g \sin 15 = 5.17 N$

Maximum frictional force that can act is $\mu m g \cos 15 = 154.54 N$

So,here only $5.17 N$ of frictional force will act upwards to keep the box at rest.

So,to move the box upwards we need to apply a minimum amount of force that will be equal to the sum of maximum frictional force and downward component of the weight of the object.(as when we will try to move the object upward,frictional force will oppose the net direction of motion,so it will act downwards along the plane)

So, minimum amount of force required to move the object with constant velocity is $\left(154.54 + 5.17\right) N = 159.71 N$

So,work done = $F . s = 159.71 \cdot 7 = 1117.97 J$