Question

A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is the acceleration of the crate if the 750-N force is maintained after the crate begins to move and the coefficient of kinetic friction is 0.12? a. 1.8 m/s b. 2.5 m/s2 c. 3.0 m/s d. 3.8 m/s2

Answer 1

Answer:

The acceleration of the crate is 1.8 m/s² so the answer is a.

Explanation:

The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)

So once we got the free body diagram, we can analyze it and build our sum of forces in the x and y directions. Notice that according to the diagram, there are 4 forces to this problem, Normal (N), Weight (W), kinetic friction (fk) and the 750N force.

As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:

Starting with the sum of forces in the y-direction, we get:

Σ$F_{y}$=0

We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.

so the sum will be:

$N-W=0$

when solving for N we get that:

N=W

where W is found by multiplying the mass of the crate by the acceleration of gravity:

N=250kg*9.8m/s²

N=2450N

Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:

$f_{k}=N$μ

where μ is the kinetic friction coefficient.

So we get that the kinetic friction is:

$f_{k}=2450N*0.12$

so

$f_{k}=294$

With this information we can go ahead and find the sum of horizontal forces:

Σ$F_{x}$=ma

In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.

so the sum of forces look like this:

750N-$f_{k}$=ma

so

750N-294N=(250kg)a

when solving for a we get:

$a=\frac{759N-294N}{250kg}\\ \\a=1.8m/s^{2}$

so the crate's acceleration is 1.82m/s².