Question

At the beginning of a new school term, a

student moves a box of books by attaching a
rope to the box and pulling with a force of
F = 84.4 N at an angle of 64°, as shown in the
figure.
The acceleration of gravity is 9.8 m/s².
The box of books has a mass of 14 kg and
the coefficient of kinetic friction between the
bottom of the box and the floor is 0.3.
64°
14 kg
p=0.3
What is the acceleration of the box?
Answer in units of m/s².

Answer 1

1.22 m/s² (3 s.f.)

Step-by-step explanation:

Draw a diagram modelling the given situation (see attachment).

• F = Friction.
  Friction always acts in the opposite direction to motion (or potential motion).
• R = Normal Reaction (perpendicular to the plane).
• Weight = mg.

Given values:

• Pulling force = 84.4 N
• Mass (m) = 14 kg
• Acceleration due to gravity (g) = 9.8 m/s²
• Coefficient of friction (μ) = 0.3

As the pulling force is at an angle to the plane (ground), resolve the force into components parallel and perpendicular to the plane.

Resolving vertically (↑) to find the Normal Reaction, R:

`\begin{aligned}\implies R+82.2 \sin 64^(\circ)&=14g\\R&=14g-82.2 \sin 64^(\circ)\end{aligned}`

The frictional force takes its maximum value when an object starts to move (or is on the point of moving):

`\boxed{F_{\text{max}}= \mu R}`

where R is the Normal Reaction and μ is the coefficient of friction.

Using F = μR to find the Frictional Force, F:

`\begin{aligned}\implies F &= 0.3\left(14g-82.2 \sin 64^(\circ)\right)\\ & =0.3\left(14(9.8)-82.2 \sin 64^(\circ)\right)\\& =0.3\left(137.2-82.2 \sin 64^(\circ)\right)\\ & = 41.16-24.66\sin 64^(\circ)\end{aligned}`

Newton's second law states that the overall resultant force acting on a body is equal to the mass of the body multiplied by the body’s acceleration:

`\boxed{F_{\text{net}}=ma}`

Resolving horizontally (→) using Newton's second law of motion to find acceleration:

`\begin{aligned}\textsf{Using} \quad F_{\text{net}}&=ma\\\\\implies 82.2 \cos64^(\circ)-(41.16-24.66\sin 64^(\circ))&=14a\\17.0383694...&=14a\\a&=(17.0383694...)/(14)\\a&=1.21702638...\\a&=1.22\;\sf m/s^2\;(3\;s.f.)\end{aligned}`

Therefore, the acceleration of the box is 1.22 m/s² (3 s.f.).