Question

A 0.600-𝑘𝑔 particle has a speed of 2.00 𝑚/𝑠 at point 𝐴 and kinetic energy of 7.50 𝐽 at point 𝐵. What is a) its kinetic energy at 𝐴? b) its speed at 𝐵? c) the total work done on the particle as it moves from 𝐴 to 𝐵?

Answer 1

a)

$$KE_A=\dfrac{1}{2}mv_A^2,$$ $$KE_A=\dfrac{1}{2}\times0.6\ kg\times(2.0\ \dfrac{m}{s})^2=1.2\ J.$$

b)

$$KE_B=\dfrac{1}{2}mv_B^2,$$ $$v_B=\sqrt{\dfrac{2KE_B}{m}}=\sqrt{\dfrac{2\times7.5\ J}{0.6\ kg}}=5\ \dfrac{m}{s}.$$

c)

$$W_{tot}=\Delta KE=KE_b-KE_a,$$ $$W_{tot}=7.5\ J-1.2\ J=6.3\ J.$$