A force of F = 70 N is applied on a block of mass M = 10 Kg placed on a horizonatl surface as shown in the figure . What will be the work done by frictional force in t = 2 sec if block was initially at rest

Answers (1)

 

The mass of the body is m= 10 kg

The force applied on the body F=70N

The coefficient of kinetic friction =0.6

Since the body starts from rest, the initial velocity of body is zero.

Time at which the work is to be determined is t=2s

The acceleration produced in the body by the applied force is given by Newtons second law of motion as

a^{\prime}=\frac{F}{m}=\frac{70}{10}=7 \mathrm{m} / \mathrm{s}^{2}$
Frictional force is given as:
$f=\mu m g=0.6 \times 10 \times 9.8=58.8$

The acceleration produced by the frictional force :

a "=-\frac{58.8}{10}=-5.88 m / s^{2} 

Therefore, the total acceleration of the body:
$a^{\prime}+a^{\prime \prime}=7+(-5.88)=1.12 m / s^{2}$
The distance traveled by the body is given by the equation of motion:
s=u t+\frac{1}{2} a t^{2}$ \\ $=0+\frac{1}{2} \times 1.12 \times(2)^{2}=2.24 \ \mathrm{m}$

 Work done by the frictional force,
$W_{f}=F \cdot s=58.8 \times 2.24=131.71 J$

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