Question

A body of mass 5 Kg rets on a rough horizontal surface of coefficient of friction 0.2. The body is pulled through a distance of 10 m by a horizontal force of 25 N. Find the kinetic energy acquired by it.

(A) 200 J
(B) 150 J
(C) 100 J
(D) 50 J

Answer 1

Hint:First start with finding the resultant force on the body of mass 5 Kg given and then use the work energy theorem and find the work done on the given body which is equal to the resultant force multiplied by the distance moved by the body which is given as 10 m in the question.



Formula used:
Frictional force, ${F_k} = \mu R$
R is normal reaction on the body.
Work done, $W = {F_{net}} \times S$
Where, ${F_{net}}$ is resultant or net force
S is the distance .


Complete answer:
Normal reaction force on the body given is R =mg
$R = 5 \times 10 = 50N$
Horizontal force on the given body, ${F_H} = 25N$ (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 , ${F_{net}} = {F_H} - {F_k}$
${F_{net}} = 25 - 10 = 15N$
At last, work done will be:
$W = {F_{net}} \times S = 15 \times 10 = 150J$
$W = 150J$



Hence, the correct answer is Option(B).


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