Explanation:
We have,
Mass of a particle is 0.6 kg
Speed at A is 2 m/s
Kinetic energy at B is 7.5 J
(a) The kinetic energy at A is given by :
[tex]K_A=\dfrac{1}{2}mv^2\\\\K_A=\dfrac{1}{2}\times 0.6\times 2^2\\\\K_A=1.2\ J[/tex]
(b) Kinetic energy at B is given by
[tex]K_B=\dfrac{1}{2}mV^2\\\\V=\sqrt{\dfrac{2K_B}{m}} \\\\V=\sqrt{\dfrac{2\times 7.5}{0.6}} \\\\V=5\ m/s[/tex]
(c) The work done on the particle as it moves form A to B is given by work energy theorem as :
[tex]W=\dfrac{1}{2}m(V^2-v^2)\\\\W=\dfrac{1}{2}\times 0.6\times (5^2-2^2)\\\\W=6.3\ J[/tex]
