Question

A wooden block of mass m resting on a rough horizontal table (coefficient of friction = $\mu$ is pulled by a force F as shown in figure. The acceleration of the block moving horizontally is

• A. $\frac{Fcos\theta }{m}$
• B. $\frac{\mu Fsin\theta }{M}$
• C. $\frac{F}{m}(cos\theta +\mu sin\theta )-\mu g$
• D. None

Answer 1

The correct option is C $\frac{F}{m}(cos\theta +\mu sin\theta )-\mu g$

Answer is C.

Let us consider a block of mass m placed on rough horizontal surface. The co-efficient of static friction between the block and suface is $\mu$. Let a pull force F be applied at an angle $\theta$ with the horizontal.
Let the pull be along the y axis. Then we have,
$\sum F_y=0$
Therefore, $N=mg-Fsin\theta$
To just move the block along the x axis, we have $Fcos\theta =\mu N=\mu (mg-Fsin\theta )$.
Rearranging the equation we get, $\frac{F}{m}(cos\theta +\mu sin\theta )-\mu g=0$.
Hence, The acceleration of the block moving horizontally is $\frac{F}{m}(cos\theta +\mu sin\theta )-\mu g=0$.