Non-consecutive forces with block and string

[BrainMan]

Homework Statement

a 4-kg block is attached to a string 2 m in length rotates on a horizontal surface. (a) If the co-efficient of friction between the block and the surface is .25, find the work done by the force of friction in each revolution of the block.

Homework Equations

Wnc= (KEf-KEi)+ (PEf-PEi)
KEi+PEi=KEf+PEf

The Attempt at a Solution

I attempted to find the work done by friction by multiplying the co-efficient of friction by the weight of the object and then multiplying that by the total distance around the circle. Unfortunately I couldn't find the distance around the circle based on the information provided. The answer to this problem was -123 J

 

[paisiello2]

Is the block tied to a string that is pinned to the circle's center?

 

[BrainMan]

I'm not sure

 

[BrainMan]

It works that way so I get it now. Thanks!

 

[HalfLostAndSearching]

.

I would suggest that to find the work done by friction in each revolution of the block, we need to first calculate the total distance traveled by the block in one revolution. This can be done by using the formula for circumference of a circle, which is 2πr, where r is the radius of the circle. In this case, the radius would be equal to the length of the string, which is 2m.

Once we have calculated the distance traveled in one revolution, we can then use the formula for work, which is W = Fd, where W is work, F is the force applied and d is the distance traveled. In this case, the force applied would be the frictional force, which can be calculated by multiplying the co-efficient of friction (0.25) by the weight of the block (4 kg x 9.8 m/s^2 = 39.2 N).

Therefore, the work done by friction in one revolution would be W = (0.25 x 39.2 N) x (2π x 2m) = 49.1 J.

Since the question asks for the work done in each revolution, we can conclude that the work done by friction in each revolution of the block would be 49.1 J.