Question

A teacher pulls a box of books on a horizontal floor with a force of 207 N directed 46.7° above the horizontal. The coefficient of kinetic friction between the box and the floor is μk = 0.222. The mass of the box and books is 43.7 kg. a. Draw a free-body diagram for the box. b. What is the normal force that the floor exerts on the box? c. Calculate the acceleration of the box.

Answer 1

Explanation:

Given that,

Mass of the box an books = 43.7 kg

Coefficient of kinetic friction = 0.222

Force = 207 N

Angle = 46.7°

(A). we draw the free body diagram

(B). We need to calculate the normal force

Using formula of normal force

∑y=0

$N+f\sin\theta=mg$

$N=mg-f\sin\theta$

Put the value into the formula

$N=43.7\times9.8-207\sin46.7$

$N=277.61\ N$

The normal force that the floor exerts on the box is 277.61 N.

(c). We need to calculate the acceleration of the box

Using formula of acceleration

$\Sigma F_{x}=ma$

$F\cos\theta-f_{k}=ma$

$a=\dfrac{F\cos\theta-\mu_{k}N}{m}$

Put the value into the formula

$a=\dfrac{207\cos46.7-0.222\times277.61}{43.7}$

$a=1.84\ m/s^2$

The acceleration of the box is 1.84 m/s².

Hence, This is the required solution.