The Work of Friction: Explained in .32m

[aqryus]

Homework Statement

A student uses a compressed spring of force constant 22N/m to shoot a 7.5 x 10^-3 kg eraser. The magnitude of the frictional force is 4.2 x 10^-2 N. How far will the eraser slide if the spring is initially compressed 3.5 cm? Use conservation of energy.

Relevant Equations

W = Fd
FE = kx
EPE = 1/2kx^2

The answer is .32m. I set the elastic potential energy as equal to the work, but at first I put the force in the work equation as (F elastic - F kinetic friction) times distance and rearranged.

1/2kx^2 = (kx-Ff) d
(0.5) (22) (0.035)^2 = (22 x 0.035-0.042) d
0.013475= 0.728 d
0.013475/0.728 = d
d = .185 m which is wrong

so then i tried with just the frictional force for the work equation

(0.5) (22) (0.035)^2 = (0.042) d
0.013475 = (0.042) d
0.013475/0.042 = d
d = .32 m which is right

So I'm confused how the work done by friction (0.042 x 0.32 = 0.013475J) can stop the applied force (22 x 0.035 x 0.32 = .2464J) if it is much lower. I thought that the work of friction had to be equal to the applied force work to make an object stop. What happens to all the energy left over? Why dont I use the total work in my equation? Thank you

 

[Lnewqban]

Welcome, @aqryus !
The spring was compressed only 3.5 cm.
Its force was reduced to zero at the end of that distance.

After that point, the eraser only had the kinetic energy that it aquired during that short distance in order to overcome the resistance of friction over a longer distance.

 

[aqryus]

Okay so after the eraser is shot the applied force is zero and only friction force does work on it right?

 

[gmax137]

Okay so after the eraser is shot the applied force is zero and only friction force does work on it right?

the applied force is zero after the spring has finished expanding, ie, after the eraser is no longer in contact with the spring.

simple drawing(s) or sketch(es) of the situation(s) can be very helpful in clarifying what is going on

 

[hutchphd]

And the friction force will do work during the short acceleration also. But work/energy conservation will give you the reported answer for the total spring-induced displacement of the eraser.

 

[kuruman]

I thought that the work of friction had to be equal to the applied force work to make an object stop.

Think again. If I push a book with my hand across a table top with a constant applied force equal to the force of friction, the book will move at constant velocity and will not stop. The work done by my hand on the book over a certain distance is equal and opposite to the work done by friction on the book. No kinetic energy is gained or lost and the book keeps moving. When I remove my hand, whatever kinetic energy the book has is dissipated as work done by friction and converted to heat until the book stops.

 

[Lnewqban]

Okay so after the eraser is shot the applied force is zero and only friction force does work on it right?

Correct.
While variable spring force is applied onto it, the eraser is gaining momentum.
Once the contact is lost, it keeps the gained momentum or gradually lose it, as in this case.

Rifles push bullets for longer distance and time than revolvers or pistols do; therefore, bullets reach a higher velocity and can cover greater distances.

 

[haruspex]

1/2kx^2 = (kx-Ff) d

There are four errors in that equation.
First, as has been pointed out above, the spring force only acts over 3.5cm, not the whole of 'd'.
Secondly, the spring force varies, so you cannot just multiply kx by a distance.
Thirdly, you are counting the spring force twice over, but in opposing senses. On the left you have all the work done by the spring force over its full expansion, x, then on the right you have it again, as though work is being done against the spring.
Fourthly, the sign of the work done against friction is wrong.

The basic reasoning is work released by spring = work done against friction, i.e. $\frac 12 kx^2=F_fd$.

 

[jbriggs444]

While variable spring force is applied onto it, the eraser is gaining momentum.

There will be a brief interval when the spring has nearly reached its relaxed state where the spring force no longer exceeds the frictional force. During this interval, the eraser will already be losing momentum despite still being in contact with the spring.

This likely has no effect on the problem solution. However, if one obtains an equation by implicitly assuming that the spring does reach the relaxed state (i.e. that the eraser slides more than 3.5 cm so that the spring's potential energy is completely used up) and solves that equation then it would be wise to check the result that is obtained to make sure that it does in fact exceed 3.5 cm.