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 in horizontal direction, the acceleration of the block will be then $(g = 10m/{s^2})$ .

A. $6.00m/{\operatorname{s} ^2}$
B. $8.0m/{s^2}$
C. $3.17m/{s^2}$
D. $10.0m/{s^2}$

Answer 1

Hint:First start with finding the resultant force on the block of mass 5 Kg given and then use the Newton’s law for finding acceleration and then put the value of the resultant force on the block and the value of the mass of the body and get the required answer for the acceleration of the block.



Formula used:
1. Frictional force, ${F_k} = \mu R$
R is normal reaction on the body.
2. R=mg
Where, m is mass of the body
3. $\Sigma F = ma$
Where, F is the net force.
And a is acceleration.



Complete answer:
Normal reaction force on the body given is R =mg
$R = 5 \times 10 = 50N$
Horizontal force on the given body, ${F_H} = 40N$ (given)
Frictional force acting on the given body, ${F_k} = \mu R = 0.2 \times 50 = 10N$
Now, the net or resultant force on the body , ${F_{net}} = {F_H} - {F_k}$
${F_{net}} = 30N$
Now, using Newton’s law;
$\Sigma F = {F_{net}} = ma$
$30 = 5 \times a$
$a = 6m/{s^2}$

Hence, the correct answer is Option(A).


Note: Be careful about the vertical and horizontal component of all the forces acting on the block and find the value of the resultant or net force acting on the given block accordingly. Also check all the units of all the quantities given before putting the values in the required equation.