Concept help. Friction, Work, Kinetic Energy

[MRGE]

I've been debating with myself on a certain question.

For example,

Two blocks. Block A has mass M and the other Block was .5M or M/2 is pushed from rest with force F at a constant rate. (Let's say a hand pushing them for a period of time). There is friction on the surface. Block B travels further than Block A

Which block has a greater net Work acting on it?

Since the Change in Kinetic Energy is equal to Work, so

1/2mv^2 = F(displacement)

Since both blocks share the same Force acting on it, I think they both will have equal Work because even though block B will have less mass than block A, it's increased velocity will compensate. So basically, the Concept here to me is if you put the same amount of Force in two blocks with different mass, you will get the same amount of work done.

But than I also thought of,

Since Block B will travel further, and Force is the same for both blocks, than W = F(displacement) This will mean there is more work done on block B than on A.

They both make sense to me, a little help?

 

[PhanthomJay]

I've been debating with myself on a certain question.

For example,

Two blocks. Block A has mass M and the other Block was .5M or M/2 is pushed from rest with force F at a constant rate. (Let's say a hand pushing them for a period of time). There is friction on the surface. Block B travels further than Block A

Which block has a greater net Work acting on it?

Since the Change in Kinetic Energy is equal to Work, so

1/2mv^2 = F(displacement)

Since both blocks share the same Force acting on it, I think they both will have equal Work because even though block B will have less mass than block A, it's increased velocity will compensate. So basically, the Concept here to me is if you put the same amount of Force in two blocks with different mass, you will get the same amount of work done.

its increased velocity will more than compensate

But than I also thought of,

Since Block B will travel further, and Force is the same for both blocks, than W = F(displacement) This will mean there is more work done on block B than on A.

They both make sense to me, a little help?

Since you have apparently determined that B travels further under the same force applied over the same time, this one makes more sense.

 

[Mindscrape]

I think we need a bit more of information towards what you're proposing. Are you trying to describe a situation in which two blocks are pushed with an equal force over an equal time (say on a non friction surface) and then released (on a friction surface), what is the work done by the frictional surface?

If, for example, you want to say two brick are released with equal velocities on a surface then the answer is pretty immediate. First using
W=F*d
W=F*.5*v0^2/a
a=F/m
W=.5*m*v0^2
which is the same as we get from work energy theorem
W=∆KE=.5*m*v0^2

So in my example the more massive block takes more work to stop.