# Work done by friction in a system

• Angeee
In summary, the work done by friction is -36.75 J as big M slides through a distance of 3 meters with a coefficient of friction of 0.25. The speed of small m when it hits the ground is 2.51 m/s. The equation used to find this solution is mk*mgd=mgh2+1/2mv^2-mgh1, which must include the kinetic energy of M and distinguish between m and M. It is also important to take the square root of the final answer to get the correct velocity.

## Homework Statement

work done by friction=? as big M slides through a distance of 3 meters.
what is the speed of small m when it hits the ground?
M=5kg m=2kg m is 3meters above ground muk is.25

I can't figure out how to post a drawing (first post)Big M is horizonatal and frictionless b4 muk attached by a string over a pulleyattached by same string to m which is the 3 meters above ground.
2.
Relevant equations

wf = -fkd wf=e2-e1

## The Attempt at a Solution

do i use cos180 since it is horizontal and I am lost on part 2

#### Attachments

• HELP.JPG
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A picture would be helpful. One could created a gif or bmp file and attach it, or put the image on a site like imageshack or flickr and use tags (roll the cursor over the buttons just above the posting box).

So M is sliding on the horizontal in one direction, and m is pulled over it in the opposite direction at some point. They are connected by a string which passes through/around a frictionless pulley. Is M sliding on a frictionless surface? So then the only friction interaction is between M and m?

If the only friction is between M and m, then one only needs to consider the friction while they are in contact. What is the length of M, i.e. how far does m travel over M?

Ok so far i have work done by friction -36.75 which the cos 180 just gives me the negative.

and the speed of small m=6.1 m/s
using the eq-----mk*mgd=mgh2+1/2mv^2-mgh1
am I on the right track?
I feel like i have missed some steps?

The work done by friction is correct. 36.75 J is lost due to friction.

The velocity seems a bit high.

See if this helps - http://hyperphysics.phy-astr.gsu.edu/hbase/hpul2.html#c1

In using the eq-----mk*mgd=mgh2+1/2mv^2-mgh1, one must include the kinetic energy of M, and don't forget to distinguish between m and M.

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Thank you-I like the link you sent. Looks like I need to still take the square root of the 6.3 which makes the correct answer...2.51m/s.

## 1. What is work done by friction?

Work done by friction is the energy expenditure required to overcome the resistance of a surface or material when it is in contact with another surface and in relative motion.

## 2. How is the work done by friction calculated?

The work done by friction can be calculated by multiplying the force of friction by the distance over which the force acts.

## 3. What factors affect the amount of work done by friction?

The amount of work done by friction is affected by the type of surfaces in contact, the force of friction, the speed of movement, and the duration of the motion.

## 4. What are the units of work done by friction?

The units of work done by friction are joules (J) in the SI system and foot-pounds (ft-lb) in the imperial system.

## 5. How does the work done by friction affect the efficiency of a system?

The work done by friction reduces the efficiency of a system by converting some of the energy into heat, which is a form of energy that cannot be used to perform work.