Question

A horizontal force of 129.4 N is applied on a 10 Kg block which rests on a horizontal surface. If the coefficient of friction is 0.3, the acceleration should be
A. $9.8\,m/{s^2}$
B. $10\,m/{s^2}$
C. $12.6\,m/{s^2}$
D. $19.6\,m/{s^2}$

Answer 1

Hint: Here we first need to find the frictional force. For this first we have to find the horizontal and the vertical component of the force given. Also the horizontal force is given. Then equate the horizontal component of the force with the frictional force in order to get the value of acceleration.

Formula used:
1. Reaction force, $R = mg$
2. Frictional force, $f = \mu R$
Where, $\mu $is the coefficient of frictional force applied.
3. Force, $F = ma$
Where, a is the acceleration of the body given in question.
And m is the mass of the block here.

Complete step by step solution:
Given information with the question:
Mass of the block given is M = 10 Kg
Coefficient of friction is $\mu = 0.3$
Force applied on the block is F = 129.4 N
Resultant force of all the forces applied on the block will be:
$F - f = ma$
Putting all the values from the question given;
$129.4 - (0.3) \times 10 \times 9.8 = 10 \times a$\
By solving, we get the value of acceleration as follows:
$\therefore a = 10\,m/{s^2}$

Hence, the correct answer is option B.

Note: The mass of the body will act in a downward direction that is reactional force is equal to mg. The force F = 129.4 N is applied on the body in a horizontal direction and frictional force is equal to $\mu R$. The resultant force will be equal to the ma and hence putting all the value we get value for acceleration of the body that is a.