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Studentka2010 [4]
2 years ago
6

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 fricti

on?
Physics
2 answers:
Papessa [141]2 years ago
7 0

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

konstantin123 [22]2 years ago
5 0

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.

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Answer: W = 0.3853 J, e = 0.052 m

Explanation: Given that,

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Answer:

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