A small block with mass 0.0450kg slides in a vertical circle of radius 0.575m on
ID: 1265133 • Letter: A
Question
A small block with mass 0.0450kg slides in a vertical circle of radius 0.575m 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 3.85N . 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 0.685N .
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
At bottom , normal force acts up , gravity force acts down.
The force equation for the block at the bottom of the circle can be given as ::
Fn - mg = mv2 /r , mv2 /r = centripetal force
Fn= normal force in up direction = 3.85 N
m = mass = 0.045 kg
r = radius = 0.575 m
v = speed at the bottom.
inserting the values
3.85 - (0.045)(9.8) = m v2 / 0.575
m v2 = 1.96
1/2 m v2 = 1.96/2 (multiplying both side by 1/2)
Kinetic energy at Bottom = 0.98 J
At the Top of circle , normal force and force of gravity acts down.
The force equation for the block at the Top of the circle can be given as ::
mg + Fn= mv2 /r
(0.045) (9.8) + 0.685 = mv2 /0.575
m v2 = 0.6475
1/2 m v2 = 0.6475/2 (multiplying both side by 1/2)
Kinetic energy at Top = 0.324 J
radius of the circle = r = 0.575 m
diameter = 2r = 2 x 0.575 = 1.15 m
the block is raised to a height equal to the diameter of circle as the block moves from bottom to top
so potential energy gained = mgh = (0.045) (9.8) (1.15) = 0.507 J
Total energy at bottom = kinetic energy = 0.98 J
Total energy at the top = kinetic energy + Potential energy = 0.324 J + 0.507 J = 0.831 J
energy lost due to friction = Total energy at bottom - Total energy at the top
energy lost due to friction = 0.98 - 0.831 = 0.149 J