A small block with mass 0.035 kg slides in a vertical circle of radius 0.425 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 4.05 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 0.660 N.
How much work was done on the block by friction during the motion of the block from point A to point B?
Okay, so I started off by doing N+mg=m(v^2)/r to get the velocity at point A. I plugged in my variables like this: 0.0350(v^2/.425)+9.8*.035=.660+.343
.0350(v^2/.425)=.660
v=2.831 m/s
Am I on the right track? I tried doing a similar thing for point B and got v=6.709 m/s. Once I got those two I plugged them into K1+U1+W=K2+mgh but I have no idea if I'm doing all of this right. Any help/confirmation would be much appreciated!
Yes, you're on the right track! Let's go step by step.
First, you correctly used the equation N + mg = mv^2/r to find the velocity at point A. Plugging in the given values, we get:
4.05 + (0.035)(9.8) = (0.035)(v^2)/0.425
Simplifying, we have:
v^2 = (4.05 + (0.035)(9.8))*0.425/0.035
v^2 ≈ 0.508
Taking the square root, we find that v ≈ 2.831 m/s (rounded to three decimal places). So, your calculation for point A is correct.
Now, let's do a similar calculation for point B.
Using the equation N + mg = mv^2/r again, we have:
0.660 + (0.035)(9.8) = (0.035)(v^2)/0.425
Simplifying, we get:
v^2 = (0.660 + (0.035)(9.8))*0.425/0.035
v^2 ≈ 8.956
Taking the square root, we find that v ≈ 6.709 m/s (rounded to three decimal places). So, your calculation for point B is also correct.
Now, let's calculate the work done by friction from point A to point B using the work-energy principle.
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. Therefore, we have:
Work by friction = K2 - K1
Since the block starts from rest at A (v1 = 0), K1 = 0. The final kinetic energy at point B is given by:
K2 = (1/2)mv^2
K2 = (1/2)(0.035)(6.709^2)
K2 ≈ 0.823 J (rounded to three decimal places)
So, the work done by friction is:
Work by friction = K2 - K1
Work by friction ≈ 0.823 J - 0
Work by friction ≈ 0.823 J
Therefore, the work done on the block by friction during the motion from point A to point B is approximately 0.823 Joules.