Question

A 0.2 kg block sliding on a horizontal table slows down from 25 m/s to 20 m/s. How much energy does the block lose due to friction?

Answer 1

Answer:

the kinetic energy lost due to friction is 22.5 J

Explanation:

Given;

mass of the block, m = 0.2 kg

initial velocity of the block, u = 25 m/s

final velocity of the block, v = 20 m/s

The kinetic energy lost due to friction is calculated as;

$\Delta K.E= K.E_f - K.E_i\\\\\Delta K.E= \frac{1}{2}mv^2 - \frac{1}{2}mu^2\\\\\Delta K.E= \frac{1}{2}m(v^2 -u^2)\\\\\Delta K.E= \frac{1}{2} \times 0.2 (20^2 - 25^2)\\\\\Delta K.E= -22.5 \ J$

Therefore, the kinetic energy lost due to friction is 22.5 J

Answer 2

Answer:

22.5 J

Explanation:

It loses the difference in kinetic energy between the starting and ending speeds and that is (1/2)(M)(V1)^2 - (1/2)(M)(V2)^2 = (1/2)(M)((V1)^2 -(V2)^2) = (1/2)(0.2 kg)((25)^2 - (20)^2) = 22.5 Joules.