You are pushing a large box across a frictionless.....(IN THE DESCRIPTION)

Question

You are pushing a large box across a frictionless floor by applying a constant horizontal force. If the box starts at rest, you have to do work W_1 in order for the box to travel a distance d in time t . How much work would you have to do, in terms of W_1, to make the box go the same distance in half the time?

PLEASE SHOW ALL WORK!!!

Answer 1

Assuming a level surface, the vertical forces of gravity and the normal force do NO work as the box moves horizontally. Also the force of gravity and the normal force cancel each other out as the acceleration is horizontal. This means that the net force, ∑F = F_1.

W₁ = F₁ d = ∑F₁ d Since ∑F₁ = m a₁, we can combine the two equations and get W₁ = m a₁ d

Now the mass and the distance are not changing in this question/scenario but if the time to travel d is cut in half it, means that the new acceleration a₂ must be larger and so must the new applied force, and net force which we will call F_{2 and} ∑F₂.

For this question, the acceleration is constant for either scenario , so we can relate the distance d to the acceleration and the time by the kinematic equation d = 1/2 at². Remember that the initial velocity is zero. Since for either scenario d is the same, this equation implies that if the time cuts in half, the new acceleration must be four times larger in scenario two versus scenario one.

Therefore W₂ = F₂*d = m* a₂* d = m *4a₁ * d But in scenario one, W₁ = m * a₁ * d

So we can see that W₂ = 4 W₁ As the time cuts in helf, the amount of work must quadruple.

Another way to reason this out is using the work-energy theorem. This says that the new work equals the change in kinetic energy.

Since the only force doing work is F₁ or F₂, the work done (W) by the force (F) equals the work done by the net force, ∑W.

For either scenario W = ∑W = ΔKE = KE_f - KE_i Since the initial velocity is zero, KE_i is zero, so W = KE_f Remember the equation for kinetic energy is KE = 1/2 mv² , so KE_f = 1/2 m v_f²

Combining equations gives W = 1/2 m v_f²

So we have to see how the final velocity changes to see how W, the work changes with scenarios.

There are a few ways to do this:

We already saw previously that the acceleration for scenario 2 was four times bigger than for scenario one. If we look at the kinematic equation v_f² = v_i² + 2ad, where v_i = 0, we can see that quadrupling 'a' means that v_f² quadruples. Remember that d is the same for both scenes. Since W = 1/2 m v_f² , then when v_f² quadruples so does W.

Another way to see this is from v_f = vi + at. When t cuts in half, a quadruples as we saw before. This means that the product of a and t must double. Since v_f = at when v_i = 0, we can see that v_f must double from scene one to scene two as "at" doubles. When v_f doubles, v_f² must quadruple.

Can we see that the final velocity must double without thinking about how the acceleration changes? Yes!

For constant acceleration: v_avg = (v_f + vi) /2 Since the box starts from rest, v_avg = v_f / 2

In general d = v_avg*t If the box covers the same distance in half the time, then the average velocity must double and since v_avg = v_f / 2 , the final velocity must also double.

This answer is very long as I don't know exactly who my audience is here. If I knew your background, this could be a much shorter answer!!