Question

A body of mass 4 kg is moving with momentum 8 kg m/s. A force of 0.2 N acts on it in the direction of motion of the body for 10 s. The increase in kinetic energy in joules is??

Answer 1

mu= 8
u=2
a=F/m=.2/4=.05
v=u+at
so v=2+.05*10=2.5
increase in kinetic energy=1/2*m(v*v-u*u)
=1/2*4(6.25-4)=4.5

Answer 2

Increase in kinetic energy = 4.5 J

• Given

          mass(m) = 4 kg

          momentum(p) = 8 kgm/s

           mu = 8                                  [∵ p = mu]

            4u = 8

            u = 2 m/s (initial velocity)

• Initial kinetic energy($k_{1}$) = $\frac{1}{2}mu^{2}$

                                               = (1/2)(4)(4)

                                               = 8 J

• Also,

            Force(F) = 0.2 N

              ma = 0.2                             [∵ F = ma]

              4a = 0.2

                 a = 0.05 $\frac{m}{s^{2} }$

• Now,

        Final velocity, v = u + at

                                   = 2 + (0.05)(10)

                                   = 2.5 m/s

• Final kinetic energy($k_{2}$) = $\frac{1}{2}mv^{2}$  

                                               = (1/2)(4)(6.25)

                                               = 12.5 J

• Change in kinetic energy = 12.5 - 8 = 4.5 J