Question

A 900 N crate is being pulled across a level floor by a force F of 340 N at an angle of 23° above the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.25. Find the magnitude of the acceleration of the crate.

Answer 1

0.958891203 m/s²

Step-by-step explanation:

N = Weight of crate = 900 N

`\mu` = Coefficient of friction = 0.25

Force of friction acting on the force applied

`f=\mu N\\\Rightarrow f=0.25* 900\\\Rightarrow f=225\ N`

Force used to pull the crate

`F=340cos 23\\\Rightarrow F=312.97165\ N`

The net force is

`F_n=F-f\\\Rightarrow F_n=312.97167-225\\\Rightarrow F_n=87.97167\ N`

Acceleration is given by

`a=(F_n)/(m)\\\Rightarrow a=(87.97167)/((900)/(9.81))\\\Rightarrow a=0.958891203\ m/s^2`

The magnitude of the acceleration of the crate is 0.958891203 m/s²