Calculate total energy expended in the acceleration

[JavaJoeUK2011]

1. Helo all,
A horizontal force of 80N acts on a mass of Kg resting on a horizontal surface. The mass is initially at rest and covers a distance of 5m in 0.92s under the action of the force. Assuming there are no energy losses due to air resistance and therefore that the acceleration is constant



2. F=ma v=s/t P=Fv



3. F=80, m=6, a=13.33,

v= 5 / 0.92 = 5.4347

P = 80 x 5.4347 = 434.776

Does this give me the total energy expended?

Thanks in advance

 

[gneill]

The problem stated that there were no energy losses due to air resistance, but it didn't mention friction with the surface. In fact, if you take the given values for the mass, force, and time and assuming no friction, the distance covered does not match given value.

Can you think of another way to calculate the energy expended (work done) by the force?

 

[JavaJoeUK2011]

would w=mas give me it
=6 x 13.33 x 5 = 399.9?

 

[gneill]

That's close. But I don't think you can count on that value for acceleration because you don't know what the frictional force is. Do a check; If you take acceleration to be f/m = 13.33 m/s², what will be the distance traveled in the given time 0.92s?

d = (1/2)at² = (1/2)(13.33m/s²)(0.92s)² = 5.64m

That does not match the given distance which is 5m, so you know there must be some other force acting and that the acceleration is not 13.33 m/s².

There's another relationship for work that does not depend upon the acceleration, velocity, or time. If you have a constant force f acting in the direction of motion over distance d, then the work done by that force is w = f*d.

 

[JavaJoeUK2011]

So if i rearrange the formula

d = (1/2)at2 = (1/2)(13.33m/s2)(0.92s)2 = 5.64m

to give me
d = (1/2)at2 = (1/2)(x m/s2)(0.92s)2 = 5m

x = 6250/529 = 11.81474 m/s2

d = (1/2)at2 = (1/2)(11.81474m/s2)(0.92s)2 = 5m

would that also give me the correct answer?

 

[gneill]

No. You only want the work done by the given force. You don't want the work from other forces acting. By contriving to find the real acceleration, you're factoring in any other forces that may be acting.

Use the f*d formula to find the work done by f alone.

 

[JavaJoeUK2011]

Do you mean w=fd = 80=f x 5 = 400

Is that correct?

 

[gneill]

Yes.

 

[HallsofIvy]

Assuming no friction forces, then you have
$v= at$ and $s= (1/2)at^2$
where a is the constant acceleration, t is the time, v is the speed at time t, and s is the distance moved in time t.

You are told that the object moved 5 m in .92 s so, from the second equation,
$5= (1/2)a(.92)^2= 0.4232a$
and so [itex]a= 5/.4232= 11.81 m/s^2[/math]

Now, you can put that into the first equation to determine the velocity at that time, then calculate the kinetic energy. The work done will be the difference between kinetic energies at start and end and the kinetic energy at the start was 0.

That is, as said, assuming no friction forces.

 

[JavaJoeUK2011]

So velocity would be v = 11.81 x 0.92 = 10.8652

so acceleration would be a = v-u/t = 10.8652-0 / 0.92 = 11.81