Question

A wooden block of mass M resting on a rough horizontal floor is pulled with a force F at an angle $\varphi$ with the horizontal. If $\mu$ is the coefficient of kinetic friction between the block and the surface, then the acceleration of the block is:

• A. $\frac{F}{M}(cos\varphi -\mu sin\varphi )-\mu g$
• B. $\frac{\mu F}{M}cos\varphi$
• C. $\frac{F}{M}(cos\varphi +\mu sin\varphi )-\mu g$
• D. $\frac{F}{M}sin\varphi$

Answer 1

The correct option is A $\frac{F}{M}(cos\varphi -\mu sin\varphi )-\mu g$

In vertical direction

$N+Fsin\varphi =Mg$

$N=Mg-Fsin\varphi$

Friction force on block

$f=\mu N$

$f=\mu (mg-Fsin\varphi )$

Acceleration of block =$(Fcos\varphi -f)/M$

$a=\frac{F}{M}(cos\varphi -\mu sin\varphi )-\mu g$