Question

A small block with mass 0.0475 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 3.85 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.

Answer #1

**at the bottom**

**Nbot - m*g = m*vbot^2/r**

**3.85 - (0.0475*9.8) = 0.0475*vbot^2/0.55**

**vbot^2 = 6.26 m/s**

**at the top**

**Ntop + m*g = m*vbot^2/r**

**0.695 + (0.0475*9.8) = 0.0475*vtop^2/0.55**

**vtop^2 = 3.66 m/s**

**from work energy relation**

**total work done = change in KE**

**Wg + Wf = dKE**

**Wg = UA - UB**

**uA = 0**

**UB = m*g*h = m*g*2r**

**-m*g*2r + Wf = (1/2)*m*(vtop^2 - vbot^2)**

**-0.0475*9.8*2*0.55 + Wf =
(1/2)*0.0475*(3.66^2-6.26^2)**

**work done by friction Wf = -0.1 J**

DONE please check the answer. any doubts post in comment box

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