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 #?

Question

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  #?

Isabella Alvarez · Accepted Answer

Inclination of #(11 pi)/12=165 degrees# means with the other direction of the horizontal level,the inclined plane makes an angle of #(180-165)=15 degrees#

So,the component of weight of the object which tries to pull it down along the inclined plane is #mg sin 15 =5.17 N#

Maximum frictional force that can act is #mumg 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 #(154.54+5.17) N =159.71 N#

So,work done = #F.s= 159.71*7=1117.97 J#

Sebastian Acosta · Answer

undefined The work done to push the object up the ramp against kinetic friction can be calculated using the formula: $ Work = mgh - \mu_k m d \cos(\theta) $, where $ m $ is mass, $ g $ is the acceleration due to gravity, $ h $ is the vertical height of the ramp, $ \mu_k $ is the kinetic friction coefficient, $ d $ is the length of the incline, and $ \theta $ is the incline angle.

Given values:
- Mass ($ m $): 2 kg
- Incline angle ($ \theta $): $ \frac{11\pi}{12} $
- Kinetic friction coefficient ($ \mu_k $): 8
- Length of incline ($ d $): 7 m

Substitute these values into the formula to find the work done.