How much mechanical energy is lost to friction if a 22-kg box slides 7.0 m down a rough inclined plane at a constant speed of 11 m/s? The inclined plane makes an angle of 33 degrees with respect to the horizontal.

Question

Quizplus · Accepted Answer

First, we need to find the gravitational potential energy of the box at the top of the inclined plane:
PE = mgh = (22 kg)(9.8 m/s^2)(sin 33°)(7.0 m) = 776 J
Next, we need to find the kinetic energy of the box at the bottom of the incline:
KE = (1/2)mv^2 = (1/2)(22 kg)(11 m/s)^2 = 1331 J
Since the box is moving at a constant velocity, the work done by friction is equal and opposite to the work done by gravity, so the mechanical energy is conserved.
ME = PE + KE = 776 J + 1331 J = 2107 J
Finally, we can find the energy lost to friction:
Friction = ME – KE = (2107 J) – (1331 J) = 776 J 
Therefore, the answer is A) 820 J, which is closest to 776 J.

How Much Mechanical Energy Is Lost to Friction If a 22-Kg