Question

A 0.500-kg particle has a speed of 1.50 m/s at point and kinetic energy of 7.70 J at point B (a) What is its kinetic energy at (b) What is its speed at m/s (c) What is the net work done on the particle by external forces as it moves from to

Answer 1

a) the kinetic energy at A is given by

$K_A=\frac{1}{2}mv^2_A$

Substituting all the values we get

$K_A=\frac{1}{2}\times 0.500\times (1.50)^2=0.5625J$

(b) kinetic energy at B is

$K_B=\frac{1}{2}mv_B^2$

From this

$v_B=\sqrt{\frac{2K_B}{m}}=\sqrt{\frac{2\times 7.70}{0.500}}=5.55 m/s$

(c) from the work-energy theorem, the net-work done is equal to change in kinetic energy

$W=K_B-K_A=7.70-0.5625=7.1375 J$