Question

Exercise 2: A 0.6 kg particle has a speed of 2 m / s at point A and kinetic energy of 7.5 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

Step-by-step 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 :

`K_A=(1)/(2)mv^2\\\\K_A=(1)/(2)* 0.6* 2^2\\\\K_A=1.2\ J`

(b) Kinetic energy at B is given by

`K_B=(1)/(2)mV^2\\\\V=\sqrt{(2K_B)/(m)} \\\\V=\sqrt{(2* 7.5)/(0.6)} \\\\V=5\ m/s`

(c) The work done on the particle as it moves form A to B is given by work energy theorem as :

`W=(1)/(2)m(V^2-v^2)\\\\W=(1)/(2)* 0.6* (5^2-2^2)\\\\W=6.3\ J`