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rewona [7]
3 years ago
6

A 15-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 29 N. Starting from rest, the sle

d attains a speed of 3.0 m/s in 8.8 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.
Physics
1 answer:
Softa [21]3 years ago
4 0

Answer:

Explanation:

v² = u² + 2 a s

3² = 0 + 2 x a x 8.8

a = .511 m / s²

Let the frictional force acting on sled be F

F = m a

29 - F = 15 x .511

29 - F = 7.665

F = 21.335 N.

F = μ R , R = mg , μ is coefficient of kinetic friction.

F = μ x mg

μ = F / mg

= 21.335 / 15 x 9.8

= .145

coefficient of kinetic friction = .145

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A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between
Darya [45]

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

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We need to calculate the acceleration of sled

Using formula of acceleration

a = \dfrac{F}{m}

Where, F = frictional force

m = mass

Put the value into the formula

a=\dfrac{\mu mg}{m}

a=\mu g

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We need to calculate the speed of the sled

Using equation of motion

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Where, v = final velocity

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Put the value in the equation

v ^2=(5.49)^2-2\times2.244\times3.89

v=\sqrt{(5.49)^2-2\times2.244\times3.89}

v=3.56\ m/s

Hence, The speed of the sled is 3.56 m/s.

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