Question

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

• A. $6.00m/se{c}^2$
• B. $8.0m/se{c}^2$
• C. $3.17m/se{c}^2$
• D. $10.0m/se{c}^2$

Answer 1

The correct option is A $6.00m/se{c}^2$

$\text{Step 1: Draw free body diagram}$ $\text{[Ref. Fig.]}$

$\text{Step 2: Finding friction}$

From figure, there is no acceleration in vertical direction. So, Applying Newton's Law in vertical direction, we get

$N=mg$

$=50N$

Maximum value of Friction $f_{max}=\mu N=0.2\times 50=10N$

Since, $f_{max}\le F$

So, sliding will happen, and kinetic friction will act, i.e.

$f_{kinetic}=\mu N=10N$

$\text{Step 3: Newton's Law for finding acceleration}$

Applying Newton's second law on block, in horizontal direction

(Taking rightward positive)

$\sum F=ma$

$\Rightarrow F-f_{kinetic}=ma$

$\Rightarrow F-\mu mg=ma$

$\Rightarrow 40-0.2\times 5\times 10=5a$

$\Rightarrow a=6m/s^2$

Hence acceleration of the block will be $6m/s^2$

Hence, Option $A$ is correct.