Question

Suppose you push a hockey puck of mass m across frictionless ice for a time t, starting from rest, giving the puck speed v after traveling distance d. If you repeat the experiment with a puck of mass 2, How long will you have to push for the puck to reach the same speed v?

Answer 1

A block pushed along the floor with velocity v slides a distance d after the pushing force is removed. If the mass of the block is doubled but its initial velocity is not changed, what distance does the block slide before stopping?

Answer 2

Please post different questions separately.

Now, as for the first question, we wish to find t in terms of v, a, and m. Since we are "pushing" the puck, we exert a force F=ma, or a=F/m. Thus, v=v_0+at is our best option, where v_0 is 0. So we can write t=v/a, or t=vm/F. Now, assuming we push with the same F, and we're trying to reach the same velocity v, we can only manipulate m and t. By manipulating m into 2, by multiplying it by 2/m, we also multiply t by 2/m. Which should give you your answer.

If the block is being "pushed" (force F=ma exerted) yet has a constant velocity, this means that the forward acceleration is being canceled by kinetic friction, which is dependent on the normal force, which is mg. The product of the coefficient of kinetic friction and the normal force should give us our force of friction. Now, since our normal force is doubled a la double the mass, the force of friction is also doubled--however, by dividing by double the mass, we still receive the original acceleration, which means it takes the same amount of time to slow down.