The sled, having rolled down the hill, has a speed of 2 m / s and moves further along a horizontal surface

| March 27, 2021 | education

The sled, having rolled down the hill, has a speed of 2 m / s and moves further along a horizontal surface so that it stops under the action of the friction force after 2 s. What are the friction force and the coefficient of friction of the sled against the surface on the horizontal section of the path, if their mass is 5 kg?

Data: V0 (speed of the given sleigh after rolling down the hill) = 2 m / s; V (final speed) = 0 m / s; t (sleigh braking time) = 2 s; m (sled weight) = 5 kg.

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

1) Friction force: Ffr = F = m * a = m * (V0 – V) / t = 5 * (2 – 0) / 2 = 5 N.

2) Coefficient of friction: Ffr = μ * m * g and μ = Ffr / (m * g) = 5 / (5 * 10) = 0.1.

Answer: The friction force is 5 N; the coefficient of friction is 0.1.



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