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Treefolk
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Calculating Work Done by Friction in a Circular Track (Answered)
A small block with mass 0.0425kg slides in a vertical circle of radius 0.400m 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, the magnitude of the normal force exerted on the block by the track has magnitude 4.00N. In this same revolution, when the block reaches the top of its path, the magnitude of the normal force exerted on the block has magnitude 0.670N.
W = K2 + U2 - K1 - U1
K=.5*m*v^2
U=mgh
F=ma
a=v^2 / r
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 = .0425(9.81+a)
94.1=9.81+a
a=84.3 <-- Radial acceleration
84.3 = v^2 / .4
33.72 = 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.
.67 = .0425(a-9.81)
15.76 = a-9.81
a = 25.57
v^2 = 25.57(.4)
v^2 = 10.228
Now plugging this into my Work equation (W = (.5mv^2 + mgh - .5mv^2))
W = .0425(.5(10.228) + (9.81*.8) - .5(33.72))
W = .0425(5.114 + 3.924 - 16.86)
W = -.33 J <-- Now this sign on this answer makes sense (frictional force would be negative), but the online homework spat it back as a wrong answer. Does anyone have any insights or hints towards the proper answer?
(The issue I had was a simple work error that I made on paper.)
Homework Statement
A small block with mass 0.0425kg slides in a vertical circle of radius 0.400m 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, the magnitude of the normal force exerted on the block by the track has magnitude 4.00N. In this same revolution, when the block reaches the top of its path, the magnitude of the normal force exerted on the block has magnitude 0.670N.
Homework Equations
W = K2 + U2 - K1 - U1
K=.5*m*v^2
U=mgh
F=ma
a=v^2 / r
The Attempt at a Solution
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 = .0425(9.81+a)
94.1=9.81+a
a=84.3 <-- Radial acceleration
84.3 = v^2 / .4
33.72 = 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.
.67 = .0425(a-9.81)
15.76 = a-9.81
a = 25.57
v^2 = 25.57(.4)
v^2 = 10.228
Now plugging this into my Work equation (W = (.5mv^2 + mgh - .5mv^2))
W = .0425(.5(10.228) + (9.81*.8) - .5(33.72))
W = .0425(5.114 + 3.924 - 16.86)
W = -.33 J <-- Now this sign on this answer makes sense (frictional force would be negative), but the online homework spat it back as a wrong answer. Does anyone have any insights or hints towards the proper answer?
(The issue I had was a simple work error that I made on paper.)
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