# Homework Help: Work and energy of block of mass

1. Nov 1, 2012

### phospho

A block of mass 10kg is pulled along a straight horizontal road by a constant horizontal force of magnitude 70N in the direction of the road. The block moves in a straight line passing through two points A and B on the road, where AB = 50m. The block is modelled as a particle and the road is modelled as a rough plane. The coefficient of friction between the block and the road is 4/7.

a) Calculate the work done against friction in moving the block from A to B.

The block passes through A with a speed of 2ms^-1.

b) Find the speed of the block at B.

for a) I got:

4/7*10g = F = 56 (taking g=9.8),
56*50 = 2800J, which is correct

for b) Work done against friction = Change in KE

so 2800 = 5v^2 - 20 and going on to get v to be 23.6, which is incorrect.

I asked a friend and he said change in KE = Work done by resultant force, and finding the resultant force to be 14N:

14*50 = 5v^2 - 20
v = 12, which is correct

however when my teacher marked this he commented to use the work energy principle, and I don't understand how I can use the work energy principle here? I understand the work energy principle states the the change in total energy = work done on the particle, but the work done on the particle is 2800J, and the change in total energy is 5v^2 - 20, as I done before giving me the wrong answer.

How can I use the work energy principle here?

2. Nov 1, 2012

### phospho

After looking at wikipedia I see it is the change in total energy = work done by the resultant force, I'm sorry my book didn't state it, it just said work done against resistance.

Also, I found this: http://www.astarmathsandphysics.com/a_level_maths_notes/M2/a_level_maths_notes_m2_the_work_energy_principle.html [Broken]

Using the formula that they give above:

0.5*2^2*10 = 0.5*v^2*10 + 14*50, which will go on to give me a complex answer :s

Last edited by a moderator: May 6, 2017
3. Nov 1, 2012

### PhanthomJay

Looks like you used the work energy theorem correctly the first time, with the help of your friend. On your last attempt, you have the work done by the net resultant force on the wrong side of the equation.

Addendum: O I see you tried to use the website equation and apply it to yours. But that equation just had friction and your problem had both friction and an applied force. Be careful when trying to blindly use equations. The net work done is the change in kinetic energy.

Last edited: Nov 1, 2012
4. Nov 1, 2012

### phospho

But in the link I've given they've shown the work done on the side I done it on?

5. Nov 1, 2012

### PhanthomJay

See my edit in my last prior post.

6. Nov 1, 2012

### haruspex

change in KE = work done by resultant force = (distance resultant moved) * (size of resultant)
= (distance resultant moved) * (applied force - frictional force)
he/she wanted you to use
change in KE + (work done against friction) = work done by applied force
change in KE + (distance friction acted)*(magnitude of friction) = (distance applied force moved) * (size of applied force)
Since the applied force and the friction moved the same distance, this comes to the same as your way.

7. Nov 1, 2012

### phospho

hm ok, so what is the work energy principle exactly:

net work done = change in total energy?

also, what if the force 70N was 50N instead, then would this be correct:

loss in KE = work done by resultant force
0.5*2^2*10 - 5v^2 = -4*50?

8. Nov 1, 2012

### PhanthomJay

no, I doubt wiki said that...the net work done = change in kinetic energy. The work done by non conservative forces (like friction and applied) is the change in total energy, which is not the same equation, although it works here because no PE change is involved.