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2 years ago

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A 5 kg block moves with a constant speed of 10 ms to the right on a smooth surface where frictional forces are considered to be

negligible.
It passes through a 2.0 m rough section of the surface where friction is not negligible, and the coefficient of kinetic friction between the block and the rough section μk is 0.2.

What is the change in the kinetic energy of the block as it passes through the rough section?

2 answers:

nirvana33 [79]2 years ago

5 0

Answer:

Work done, W = 19.6 J

Explanation:

It is given that,

Mass of the block, m = 5 kg

Speed of the block, v = 10 m/s

The coefficient of kinetic friction between the block and the rough section is 0.2

Distance covered by the block, d = 2 m

As the block passes through the rough part, some of the energy gets lost and this energy is equal to the work done by the kinetic energy.

$W=\mu_kmgd$

$W=0.2\times 5\times 9.8\times 2$

W = 19.6 J

So, the change in the kinetic energy of the block as it passes through the rough section is 19.6 J. Hence, this is the required solution.

Sladkaya [172]2 years ago

3 0

Answer:

19.6 J

Explanation:

mass of block, m = 5 kg

initial velocity, u = 10 m/s

coefficient of friction, μk = 0.2

distance, s = 2 m

Let v be the velocity after covering the friction surface

use third equation of motion

v² = u² + 2as

v² = 10² - 2 x 0.2 x 9.8 x 2

v² = 100 - 7.84

v = 9.6 m/s

initial kinetic energy, ki = 0.5 x m x u²

Ki = 0.5 x 5 x 10 x 10 = 250 J

final kinetic energy

kf = 0.5 x m x v² = 0.5 x 5 x 9.6 x 9.6 = 230.4 J

Change in kinetic energy, K = Kf - Ki = 250 - 230.4 = 19.6 J

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