Question

A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40 N is applied as shown in . figure the acceleration of the block will be $(g=10m/s^2)$

• A. $5.73m/se{c}^2$
• B. $8.0m/se{c}^2$
• C. $3.17m/se{c}^2$
• D. $10m/se{c}^2$

Answer 1

The correct option is A $5.73m/se{c}^2$

Let us consider a block of mass m = 5 kg placed on rough horizontal surface. The co-efficient of static friction between the block and suface is $\mu$ = 0.2. Let a pull force F = 10 N be applied at an angle $\theta$ = 30 degrees with the horizontal. Let g be $10m/s^2$.
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, $\frac{F}{m}(cos\theta +\mu sin\theta )-\mu g=0$.
Substituting the values in the above equation, we get,
$\frac{40}{5}(cos30+0.2\times sin30)-0.2\times 10=5.73m/s^2$
Hence, The acceleration of the block moving horizontally is $5.73m/s^2$.