Question

A 0.1 kilogram block is released from rest at point A as shown, a vertical distance h the ground. It slides down an inclined track, around a circular loop of radius 0.5 meter, then up another incline that forms an angle of 30° with the horizontal. The block slides off the track with a speed of 4 m/s at point C, which is a height of 0.5 meter above the ground. Assume the entire track to be friction less and air resistance to be negligible.

a. Determine the height h . 

b. On the figure below, draw and label all the forces acting on the block when it is at point B. which is 0.5 meter above the ground.

c. Determine the magnitude of the velocity of the block when it is at point B. 

d. Determine the magnitude of the force exerted by the track on the block when it is at point B. 

e. Determine the maximum height above the ground attained by the block after it leaves the track. 

f. Another track that has the same configuration, but is NOT friction less, is used. With this track it is found that if the block is to reach point C with a speed of 4 m/s, the height h must be 2 meters. Determine the work done by the frictional force.

 

Answer 1

(a) Apply energy conservation from point A to point C setting point C as h = 0 location (note: to find h as shown in the diagram, we will have to add in the initial 0.5m below h=0 location)

(c) Since the height at B and the height at C are the same, they would have to have the same velocities v_b = 4m/s

Alternatively you can do energy conservation setting h = 0 at point C. Then K_c = U_top + K_top keeping in mind that at the top the block has a kinetic energy related to its velocity there which is the same as v_x at point C. 

(f) Since the block will have the same total energy at point C as before but it will lose energy on the track the new initial height h is larger than before. To find the loss of energy on the track, you can simply subtract the initial energies in each case. 

U_new – U_old = mgh_net - mgh_old 

(0.1)(9.8)(2-1.32) = 0.67J lost.