Question

A student moves a box of books down the hall by pulling on a rope attached to the box. The student pulls with a force of 171 N at an angle of 30.7 ◦ above the horizontal. The box has a mass of 38.9 kg, and µk between the box and the floor is 0.19. Find the acceleration of the box.

Answer 1

-2.34 ms⁻²

Explanation:

As per the question,

We have been provided the information that is

Force at an angle 30.7° = 171 N

Mass of the box = 38.9 kg

Coefficient of friction between box and the floor = 0.19

Now,

By drawing the free body diagram and taking the horizontal and vertical component for force 171 N(as show below)

`At equilibrium:

N + 171 sin 30.7° = mg ...................equation (i)

ma + 171 cos 30.7° = f ..................equation (ii)

Where,

f = frictional force

As we know that,

Frictional force is given by:

f = μN

Where,

μ = coefficient of friction

N = normal reaction

From equation (i)

N = mg - 171 sin 30.7°

∴ f = μ ( mg - 171 sin 30.7°) ............equation (iii)

From equation (ii) and (iii), we get

ma + 171 cos 30.7° = μ ( mg - 171 sin 30.7°)

Put the value of mass, μ and g = 9.8 m/s²

(38.9)a + 171 cos 30.7° = 0.19 × ( (38.9 × 9.8) - 171 sin 30.7°)

(38.9)a + 146.889 = 55.861

a = -2.34 ms⁻²

Hence, the acceleration of the box is = - 2.34 ms⁻²