A spring with a spring constant of 85 N/m is attached to a 4.0 kg block of wood. The coefficient ...

Question

A spring with a spring constant of 85 N/m is attached to a 4.0 kg block of wood. The coefficient of static friction between the block and the table is 0.50. How far must be the spring stretched for the block to just begin to move if...

a) the applied force on the spring is horizontal

b) the applied force of the spring is 25° above horizontal

c) the applied force of the spring is 15° below horizontal

Answer 1

a) The spring force is F = kx where k = 85 N/m and x is the stretch of the spring. The frictional force is F= μmg. where μ = 0.50, m = 4 kg, and g = 9/8 m/s².

a) When the applied force is horizontal, all of the applied force of the spring is applied horizontally. To move the block, we must have:

kx ≥ μmg

Solve for x.

b) When the spring is applied at 25° above horizontal, the applied spring force has a horizontal component of kx cos25° and a vertical component of -kx sin25° (directed upward). Ignoring torque, the vertical component opposes and reduces the gravitational force of the block. The new condition for moving the block is:

kx cos25° ≥ μ(mg-kx sin25°)

kx cos25° ≥ μmg - μkx sin25°

kx cos25° + μkx sin25° ≥ μmg

kx(cos25° + μ sin25°) ≥ μmg

x ≥ μmg/k(cos25° + μ sin25°)

c) This part works just like part (b) except the angle is 15° and the vertical component adds to the gravitational force on the block.