Question

A block of mass 2 Kg placed on a floor experiences an external force in horizontal direction of
20N, frictional force of 6N and normal force of 20N. The body travels a distance of 10m under
the combined effect of all these force. If Initially body is at rest then what is the kinetic energy
of the body at the end of 4m?
​

Answer 1

Kinetic energy = 28 joule

Explanation:

According to the question, we make the free body diagram. (refer the figure)

Apply Newton's law of motion;

• Net force = $mass \times acceleration$

• horizontal force - friction =$m\times a$

• $20 N - 6 N = m\times a$     ........(1)

•                   mg = 20N

       $m\times 10 = 20$

                             m = 2 kg

put this value in (1), we get.

• $14 = 2\times a$

          $a = 7 \ m/s^2$

Since body is in rest initially. So u = 0 and s = 4 meter (given)

Now, apply second law of motion, we get.

• $s = ut + \dfrac{1}{2} at^2$

        $4 = 0 + 7t^2$

     $t = \sqrt{\dfrac{4}{7}} \ seconds$

Now, to find out the final velocity at the end of 4 meter.

Apply v = u +at

• $v = 0 + 7\times\sqrt{\dfrac{4}{7}}$

        $v = \sqrt{28}$m/s

Kinetic energy = $\dfrac{1}{2}mv^2$

Kinetic energy = $\dfrac{1}{2}2\times 28$

Thus, Kinetic energy = 28 joule

Answer 2

Answer: 56 J

Explanation:

m = 2kg

F= 14N

S = 4m

u=0

F = ma

14 = 2a

a = 7 m/s²

v² = u²+ 2as

v² = 0 + 2*7*4

v² = 56

KE = 1/2 mv²

= 1/2*2*56

=56 J