Question

A bullet of mass 20g leaves a rifle at an initial with speed 100m/s and strikes a target at the same level with speed 50m/s. The amount of work done by the resistance of air??

Answer 1

Answer:

75 J

Explanation:

work done is equal to change in kinetic energy according to work energy theorem

so 1/2 x(m x v^2)-1/2(m xu^2)=w

M=.02 Kg v=100 u=50

1/2(.02 x100 x100-1/2(.02 x 50 x50=w

100-25=w

w=75 J

Answer 2

Answer:

The amount of work done by the resistance of air is 75 J.

Explanation:

Given that,

Mass of bullet = 20 g

Initial speed of bullet = 100 m/s

Final speed = 50 m/s

We need to calculate the work done

Work done = change in kinetic energy

$W=\Delta K.E$

$W=\dfrac{1}{2}mv_{i}^2-\dfrac{1}{2}mv_{f}^2$

Put the value into the formula

$W=\dfrac{1}{2}\times20\times10^{-3}\times(100)^2-\dfrac{1}{2}\times20\times10^{-3}\times(50)^2$

$W=75\ J$

Hence, The amount of work done by the resistance of air is 75 J.