e (m 23 kg) is resting on a frictionless inclined plane while suspended from a w

Question

e (m 23 kg) is resting on a frictionless inclined plane while suspended from a wall spring. The plane is inclined at an angle of 30° from the horizontal, and the spring is stretched from its equilibrium position by 25 cm. What is the spring constant of the spring? If the frictionless inclined plane is now replaced with an otherwise identical plane except with a rough surface, the crate now stretches the spring by 12 cm from its cquilibrium position. What is the coefficient of static friction? You can assume that the static friction is at its maximum value. 30

Explanation / Answer

Force acting on the crate in the downward direction along the inclined plane -

F = m*g*sin30

this force must be equal to kx, developed in the spring.

So, kx = m*g*sin30

=> k = m*g*sin30 / x

given, m = 23 kg, g = 9.81 m/s^2, x = 25 cm = 0.25 m

So, k = (23*9.81*0.5) / 0.25 = 451.26 N/m

Now when the surface is rough.

suppose 'u' is the coefficient of static friction in this case.

so the force-mass equation becomes -

m*g*sin30 = kx1 + u*m*g*cos30

put the values -

=> 23*9.81*0.50 = 451.26*0.12 + u*23*9.91*0.87

=> 112.82 = 54.15 + 196.3*u

=> u = 0.30

So, value of coefficient of static friction = 0.30

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