Question

A body of weight 64 N is pushed with just enough force to start moving across a horizontal floor and the same force continues to act afterwards. If the coefficients of static and dynamic friction are 0.6 and 0.4 respectively. The acceleration of the body will be
(acceleration due to gravity = g)
A. $\dfrac{g}{2}$
B. $0.64g$
C. $\dfrac{g}{{32}}$
D. $0.2g$

Answer 1

Hint:Here we first need to find the frictional force that will cause the body to start moving. For this first we have to find the horizontal and the vertical component of the force given. Also we need to find the reactional force here that will equal to mg. Frictional force will be equal to multiplication of mg and frictional coefficient.

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

Complete step by step solution:
Start with the given information:
Coefficient of static friction = 0.6
Coefficient of dynamic friction = 0.4
Mass of body, m = 64 N
Frictional force,
$f = \mu R = \mu mg$
Putting the value of coefficient of friction:
$f = mg(0.6 - 0.4)$
By solving;
$f = 0.2mg$ (equation 1)
Also, we know that;
$F = ma$ (equation 2)
Equating both the equation 1 and 2, we get;
$ma = 0.2mg$
$\therefore a = 0.2g$

Hence, the correct answer is option D.

Note: Here the vertical component of force balances the mg force applied on the body in downward vertical direction and the horizontal component balances the frictional force applied on the body. Reactional force will be resultant of all the vertical forces applied on the given body. Then finally multiplying it with the frictional coefficient we will get the final answer.