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A 52 N sled is pulled across a cement sidewalk at a constant speed. A horizontal force of 36 N is exerted. What is the coefficient of sliding friction between the sidewalk and the metal runners of the sled? [Friction Numerical]

Numerical Question: A 52 N sled is pulled across a cement sidewalk at a constant speed. A horizontal force of 36 N is exerted. What is the coefficient of sliding friction between the sidewalk and the metal runners of the sled?

Solution:

Weight of the sled = 52 N. Here the normal force on the Slade = FN (sled) = 52 N
Hence the friction working against its motion = FF= μFN = μ 52 = 52μ (here μ is the coefficient of sliding friction between the sidewalk and the metal runners of the sled)

As the sled is moving at a constant speed that means the net force on it is zero. This implies that the horizontal force of 36 N on it just equals the friction in magnitude.
Hence, 52μ = 36
=> μ = 36/52 = 0.69

Answer: The coefficient of sliding friction between the sidewalk and the metal runners of the sled is 0.69

See also  If the coefficient of sliding friction between a 25 kg crate and the floor is 0.45, how much force is required to move the crate at a constant velocity across the floor? [Friction Numerical]
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