A small block with mass 0.0425 k g slides in a vertical circle of radius 0.525 m

Question

A small block with mass 0.0425kg slides in a vertical circle of radius 0.525m 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 4.05N . 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 0.695N .

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

please show details to how you answered it

Thank you

Explanation / Answer

The easiest method to approach this problem appeared to be to compare the difference in energy between the two points and solve for W.

At the bottom of the circle:

N = mg + ma = m(g+a) <-- This is from the FBD I drew out, the normal force equates to gravity force + radial acceleration.

4.05 = .0425(9.81+a)

a=85.48 <-- Radial acceleration

85.48 = v^2 / .525

44.877 = v^2 <--since my later equation contains v^2 as a variable, I didn't feel I needed to take the square root.

At the top of the circle:

ma = N + mg --> N = ma - mg = m(a-g) <-- Again, FBD. Similar process to above for the work.

.695 = .0425(a-9.81)

a = 26.12

v^2 = 26.12(.525)

v^2 = 13.713

Now plugging this into my Work equation (W = (.5mv^2 + mgh - .5mv^2))

W = .0425(.5(13.713) + (9.81*1.05) - .5(44.877))

W = -.224 J <-- Now this sign on this answer makes sense (frictional force would be negative),

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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