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;
[tex]\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[/tex]
Therefore, the kinetic energy lost due to friction is 22.5 J
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.
