Question

A force of 10. N toward the right is exerted on a wooden crate initially moving to the right on a horizontal wooden floor. the crate weight 25 N.
1)calculate the magnitude of the force of friction between the crate and the floor.



2)What is the magnitude of the net force acting on the crate?



3)is the crate accelerating? explain.

Answer 1

1) 5 N

The crate is initially moving, so we must calculate the force of kinetic friction, which is given by:

$F_f = \mu_k mg$

where

$\mu_k=0.2$ is the coefficient of friction between the crate (made of wood) and the floor (made of wood). The coefficient of kinetic friction between wood and wood is about 0.2.

$mg=25 N$ is the weight of the crate

Substituting the numbers into the formula, we find

$F_f=(0.2)(25 N)=5 N$

2) 5 N

There are two forces acting on the crate along the horizontal direction:

- The force that pushes the crate toward the right, of magnitude $F=10 N$

- The force of friction, which acts in the opposite direction (so, towards the left), of magnitude $F_f = 5 N$

Since the two forces are in opposite directions, the net force is given by their difference:

$F_{net}=F-F_f = 10 N-5 N=5 N$

3) Yes

The crate is accelerating. In fact, according to Second Newton's Law:

$F_{net}=ma$ (1)

where Fnet is the net force on the crate, m is its mass, a is its acceleration. We can immediately see that since Fnet is not zero, the acceleration is also non-zero, so the crate is accelerating.

We can even calculate the magnitude of the acceleration. In fact, the mass of the crate is given by:

$m=\frac{Weight}{g}=\frac{25 N}{9.8 m/s^2}=2.55 kg$

And by using (1) we find

$a=\frac{F_{net}}{m}=\frac{5 N}{2.55 kg}=1.96 m/s^2$