# Work vs mass

## Homework Statement

Is more work done on a large or a small mass?

F = d/dt (mv)
W = Fd

## The Attempt at a Solution

Greater the mass, greater the velocity according to F = d/dt (mv)

If the problem is given that
a glider of mass is free to slide along a horizontal air track
one asks this given question, should one consider that, since motion only exists in the x-direction, so the mass is an independent clause? Does this sounds right?

I mean in momentum, when a small mass hits a larger mass, we see the smaller mass moves at a greater velocity, assuming it is elastics.

However, when I look at W = E (mechanical energy) and KE, if the system is conservated with no external force, we always cancel out m (mass).

But how come the answer is "no difference"?

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collinsmark
Homework Helper
Gold Member
Is more work done on a large or a small mass?
The problem statement is kind of vague. Is it asking

If a constant force is applied to an object over a given distance d, is more work done if the object has large or small mass?

I'm not sure if that's what the question is, but I'll assume it is for now.

F = d/dt (mv)
W = Fd

## The Attempt at a Solution

Greater the mass, greater the velocity according to F = d/dt (mv)

If the problem is given that

one asks this given question, should one consider that, since motion only exists in the x-direction, so the mass is an independent clause? Does this sounds right?
I don't think I follow you. :uhh:

But if you apply a constant force on an object for a set distance d, the work done is independent of the mass. It only depends on the force and distance.

$$W = \vec F \cdot \vec d$$

I mean in momentum, when a small mass hits a larger mass, we see the smaller mass moves at a greater velocity, assuming it is elastics.

However, when I look at W = E (mechanical energy) and KE, if the system is conservated with no external force, we always cancel out m (mass).

But how come the answer is "no difference"?
Perhaps the fact that you are able to "cancel out" m, means that the answer does not depend on m. If you push a mass, starting from rest, with a constant force F over a distance d, the object will have greater speed when it reaches the distance d if it has a smaller mass. The object would have a slower speed if it has a larger mass. But if you calculate the kinetic energies of the different objects (1/2)mv2, the result is the same for both.

You can convince yourself of this by using the kinematics equations for constant acceleration,

$$d = \frac{1}{2}at^2 +v_0t + d_0$$

$$a = \frac{v_f - v_i}{t}$$

and

$$F= ma$$

Using these equations, calculate the velocity of an object, starting from rest, that has been pushed by a constant force F (on a frictionless surface) for a distance d. Then calculate the object's kinetic energy. You'll find that the kinetic energy is not a function of mass, in this situation. As a matter of fact, you'll find that the kinetic energy is a simple function of force and distance.

PhanthomJay
Homework Helper
Gold Member

## Homework Statement

Is more work done on a large or a small mass?

... how come the answer is "no difference"?
What's the question?? You are not correctly stating work/energy relationships.