Question

A 91 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 340 N . For the first 15m the floor is frictionless, and for the next 15 m the coefficient of friction is 0.22.

What is the final speed of the crate?

Answer 1

First, we need to find the accelerations across the two different surfaces. We'll call them a1 (across frictionless floor) and a2 (across floor with friction).

To find a1, all we need is our good old friend Net F=ma

for the first part of the problem, the Normal Force of the Block is canclled out by the Force of gravity. And since we don't hav friction, the only force we have is the force of 340 N

F=ma1

a1=340/91=3.74 m/s^2

Now, let's find a2

For the 2nd part of the problem (the half with friction) we are adding a force, friction. Remember friction = Normal Force * mu (coefficent of friction).

We know that our normal force is equal to mg because they cancel out each other. So friction force = m*g*coefficient of friction.

Now, let's use F net = ma

since friction is always resisting the direction we want to go, friction is going to be negative.

so, 340- friction = m * a2

340-(0.22*91*10)=91*a2   (here friction=mue*mg)

a2=1.54 m/s^2

So let's find the velocity at the end of the first stage. We can use Vf^2 = V0^2 + 2ax

Vf^2 = (0)^2 + 2 * a1 * x

Vf=sqrt(2*3.74*15)=10.59 m/sec

Vf = 10.42 m/s, this is the ending velocity at the end of the first part, now we can use the same formula to find the final final velocity. Using the velocity we just found as the initial velocity

Vf^2 = Vo^2 + 2*a2*x

Vf= sqrt(10.59^2+2*1.54*15)=12.58 m/sec