Answer:
1.22 m/s² (3 s.f.)
Explanation:
Draw a diagram modelling the given situation (see attachment).
Given values:
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:
[tex]\begin{aligned}\implies R+82.2 \sin 64^{\circ}&=14g\\R&=14g-82.2 \sin 64^{\circ}\end{aligned}[/tex]
The frictional force takes its maximum value when an object starts to move (or is on the point of moving):
[tex]\boxed{F_{\text{max}}= \mu R}[/tex]
where R is the Normal Reaction and μ is the coefficient of friction.
Using F = μR to find the Frictional Force, F:
[tex]\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}[/tex]
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:
[tex]\boxed{F_{\text{net}}=ma}[/tex]
Resolving horizontally (→) using Newton's second law of motion to find acceleration:
[tex]\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&=\dfrac{17.0383694...}{14}\\a&=1.21702638...\\a&=1.22\;\sf m/s^2\;(3\;s.f.)\end{aligned}[/tex]
Therefore, the acceleration of the box is 1.22 m/s² (3 s.f.).
