Potential energy, momentum

Question

As shown in the figure below, a 2.25 kg block is released from rest on a ramp of height h.

When the block is released, it slides without friction to the bottom of the ramp and then continues across a surface that is frictionless except for a rough patch of width 15.0 cm that has a coefficient of kinetic friction 𝜇_k = 0.690.



Find h (in m) such that the block's speed after crossing the rough patch is 3.50 m/s. (Enter a number.)

Answer 1

The initial energy is all gravitational potential energy and is equal to "mgh":

E_{initial (grav pot)} = mgh = (2.25)(9.81)(h) = 22.0725h joules

When the block reaches the bottom of the ramp, all the initial gravitational potential energy has been converted into kinetic energy but it is still equal to 22.0725h joules.

Crossing the rough spot, there will be energy lost in work done by the frictional force. W = F_Fd where F_F is the force of friction and "d" is the distance the force is applied (15 cm or 0.15 meters). We also know that F_F = μN where μ is the kinetic coefficient of friction and "N" is the normal force. Summing the forces in the y-direction, we find that N = mg = (2.25)(9.81) = 22.0725 N so therefore F_F = (0.690)(22.0725) = 15,23 N and the work done is: W = (15.23)(0.15) = 2.2845 joules

After the block crosses the rough patch, it has kinetic energy equal to 1/2mv²:

E_{final (Kinetic)} = 1/2mv²

E_{final (Kinetic)} = (1/2)(2.25)(3.50)² = 13.78125 joules

So E_{initial (grav pot)} - W = E_{final (Kinetic)} or

22.0725h - 2.2845 = 13.78125

22.0725h = 16.0658

h = 0.728 meters