Question

A 20.0 kg box slides down a 12.0 m long incline at an angle of 30.0 degrees with the horizontal. A force of 50.0 N is applied to the box to try to pull it up the incline. The applied force makes an angle of 0.00 degrees to the incline. If the incline has a coefficient of kinetic friction of 0.100, then the increase in the kinetic energy of the box is: 300 J.

372 J.
410 J.
455 J.
525 J.

Answer 1

`KE_f = 372 J`

Step-by-step explanation:

The forces on the box while it is sliding down are

1). Component of the weight along the inclined plane

2). Friction force opposite to the motion of the box

3). Applied force on the box

now we know that component of the weight along the inclined plane is given as

`F_g = mgsin\theta`

`F_g = (20.0)(9.8)sin30 = 98 N`

Now we know that other component of the weight of object is counterbalanced by the normal force due to inclined plane

`F_n = mgcos\theta`

`F_n = (20.0)(9.8)cos30 = 170 N`

now the kinetic friction force on the box is given as

`F_k = \mu F_n`

`F_k = 0.100(170) = 17 N`

now the Net force on the box which is sliding down is given as

`F_(net) = F_g - F_k - F_(applied)`

`F_(net) = 98 - 17 - 50 = 31 N`

now the work done by net force = change in kinetic energy of the box

`F.d = KE_f - KE_i`

`31(12) = KE_f - 0`

`KE_f = 372 J`