Question

It shows a 2.0 kg block is being pushed along a rough horizontal surface by a force F=30N at an angle 60° from the normal. If the block moves at constant acceleration 0.5ms-² , what is the coefficient of friction ?

Answer 1

According to Newton's second law, the interacting forces can be written as

$$F\text{ sin}60°-f=ma.$$

The force of friction here is

$$f=F\text{ sin}60°-ma=\mu N,\\\space\\ \mu=\frac{F\text{ sin}60°-ma}{N},$$

where the normal force is

$$N-mg+F\text{ cos}60°=0,\\ N=mg-F\text{ cos}60°.$$

So, the coefficient of friction is

$$\mu=\frac{F\text{ sin}60°-ma}{mg-F\text{ cos}60°}=5.43.$$