A small block with mass 0.0300 kg slides in a vertical circle of radius 0.550 m

Question

A small block with mass 0.0300 kg slides in a vertical circle of radius 0.550 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point A, the magnitude of the normal force exerted on the block by the track has magnitude 3.75 N . In this same revolution, when the block reaches the top of its path, point B, the magnitude of the normal force exerted on the block has magnitude 0.695 N .

How much work was done on the block by friction during the motion of the block from point A to point B?

Express your answer with the appropriate units.

Explanation / Answer

We will use the concept of conservation of energy to determine the work done by friction. Plus, we also need to keep into cosideration the circular motion that the block is doing. That would imply that the net normal force acting on the block would be contributing to the centripetal force acting on the block so as to enable the circular motion.

For point A: N1 - mg = mu^2 / R

For point B: N2 + mg = mv^2 / R

We will use the above equation to determine the difference in kinetic energy of the block at two points and then conserve the energy to find the required work done.

Putting the values in the equations above, we get:

3.75 - 0.03g = 0.03u^2 / 0.550

Or 0.03u^2 = 1.901

or KE1 = 0.9503 J

Also, 0.695 + 0.03g = 0.03v^2 / 0.550

or 0.03v^2 = 0.544115

or KE2 = 0.2721 J

Therefore, we have energy equation as: KE1 = Wf + MgH + KE2

That is, 0.9503 = Wf + 0.32373 + 0.2721 = Wf + 0.59583

Or Wf = 0.35447 Joules

Therefore, the amount of work done by friction is 0.35447 Joules

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.