An object with a mass m slides down a rough 37° inclined plane where the coefficient of kinetic friction is 0.20. If the plane is 10 m long and the mass starts from rest, what will be its speed at the bottom of the plane?

Question

Physics

Jimena Le

27 January, 16:25

An object with a mass m slides down a rough 37° inclined plane where the coefficient of kinetic friction is 0.20. If the plane is 10 m long and the mass starts from rest, what will be its speed at the bottom of the plane?

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Answers (1)

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Trevon Anderson · Answer 1

Answer: 9.312 m/s

Explanation:

The friction force (opposite to the motion) is Fa = μ*m*g*cos (α) with μ = kinetic friction. The force that makes the motion is

F = m*g*sin (α).

The Newton's law gives:

F - Fa = m*a

m*g*sin (α) - μ*m*g*cos (α) = m*a

g*sin (α) - μ*g*cos (α) = a so a = 4.335 m/s²

It's a uniformly accelerated motion:

Space

S = 0.5*a*t²

10 = 0.5*a*t²

=> t = 2.148 s

Velocity

V = a*t = 9.312 m/s.