Question

A body of mass 2kg is resting on a rough surface. A force of 20N is applied for 10s parallel to the surface. If coefficient of kinetic friction between the surfaces in contact is 0.2 then calculate
Work done by the applied force and change in kinetic energy of the body in 10s

Answer 1

Answer:

A body of mass 2kg is resting on a rough surface. A force of 20N is applied for 10s parallel to the surface. If coefficient of kinetic friction between the surfaces in contact is 0.2 then calculate

Work done by the applied force and change in kinetic energy of the body in 10s

Answer 2

Answer:

A force  $20N$ is applied to a body of mass $2Kg$ is resting on a rough surface for $10s$ parallel to the surface. If the coefficient of kinetic friction is $0.2$ then work done by the applied force is $8KJ$ and the kinetic energy of the body is  $6.4J$

Explanation:

We have the formula to find work done by an applied force $F$

$W=F.d$  

From the equation of motion, the distance traveled by a body with acceleration $a$ is

$d=ut+\frac{at^{2} }{2}$

Initially, the body is at rest

⇒$u=0$

to calculate the acceleration of the body we have,

$a=\frac{F_{net} }{m}$                

Net force is

$F_{net} =F_{appl} - F_{fric}$

$F_{fric} =\mu mg$

From the question, mass $m=2Kg$ $\mu = 0.2, t=10s, g=10\frac{m}{s^{2} }$

Substituting the values we get $F_{net} =20-(0.2\times 2\times 10)=16N$

Now acceleration,  $a=\frac{16}{2} =8\frac{m}{s^{2} }$

And distance traveled, $d=0+\frac{1}{2}\times 8\times 10^2 =400m$

Then work done, $W=20\times 400=8kJ$

The kinetic energy of the body $KE=\frac{1}{2} mv^{2}$

but, $v^{2} =u^{2} +2as$ (equation of motion)

$KE=\frac{1}{2}\times 2\times 2\times 400\times 8=6\cdot 4kJ$