Answer to Question #91262 in Mechanics | Relativity for Umar

Question

A body of mass 10kg and initially at rest is subjected to force of 20N for 1s. Calculate the change in kinetic energy

Answer 1

Let's first find the acceleration of the body from the Newton's Second Law of Motion:

$$F = ma,$$ $$a = \dfrac{F}{m} = \dfrac{20 N}{10 kg} = 2 \dfrac{m}{s^2}.$$

Then, we can find the final velocity of the body from the kinematic equation:

$$v_f = v_i + at,$$

here, $v_f$ is the final velocity of the body, $v_i = 0$ is the initial velocity of the body, $a$ is the acceleration of the body and $t$ is time.

Then, we get:

$$v_f = v_i + at = 0 \dfrac{m}{s} + 2 \dfrac{m}{s^2} \cdot 1 s = 2 \dfrac{m}{s}.$$

Finally, we can find the change in kinetic energy:

$$\Delta KE = KE_f - KE_i,$$ $$\Delta KE = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}m(v_f^2 - v_i^2),$$ $$\Delta KE = \dfrac{1}{2} \cdot 10kg \cdot ((2 \dfrac{m}{s})^2 - (0 \dfrac{m}{s})^2) = 20 J.$$

Answer:

$\Delta KE = 20 J.$