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

Question

1 month 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?

Physics

2 answers:

Yuliya22 [3.3K] · Answer 1 · 2026-03-22T05:19:16+03:00

Answer:

The work done, W = 19.6 J

Explanation:

It’s provided that

The mass of the block, m = 5 kg

The velocity of the block, v = 10 m/s

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

Distance traveled by the block, d = 2 m

As the block traverses the rough section, it loses energy 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

Thus, the change in kinetic energy of the block moving through the rough section is 19.6 J. Consequently, this is the required answer.

ValentinkaMS [3.4K] · Answer 2 · 2026-03-22T05:22:59+03:00

Answer:

19.6 J

Explanation:

mass of the block, m = 5 kg

initial velocity, u = 10 m/s

friction coefficient, μk = 0.2

distance, s = 2 m

Let v represent the velocity after passing over the frictional surface

Employ the third equation of motion

v² = u² + 2as

v² = 10² - 2 x 0.2 x 9.8 x 2

v² = 100 - 7.84

Thus, 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

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