A spring (with spring constant k = 20.0 N/m) is initially compressed 50 cm. When

Question

A spring (with spring constant k = 20.0 N/m) is initially compressed 50 cm. When the spring is released, it launches a block (of mass m = 0.5 kg) up a ramp that is elevated at an angle = 30° (the spring is aimed along the ramp). The coefficients of static and kinetic friction between the block and the inclined plane are µs = 0.25 and µk=0.15, respectively. (a) What is the block’s initial velocity up the ramp (as it disengages from the spring)? (b) What is the maximum distance from the end of the spring that the block reaches? (c) How long is it sliding until it momentarily comes to rest? (d) How long until the block reaches the spring (now at its relaxed length) again. Please show all work.

Explanation / Answer

0.5 kg block is attached to a spring of force constant 20 N/m . The block is compressed 50 cm and released from rest.

Find the speed of the block as it passes through equilibrium (for the first time) if the coefficient of friction between block and surface is 0.25

Force = k * distance compressed
k = 20 N/m
distance stretched = 50 cm = 0.50 m
Force = 20*0.50 = 10 N

As the spring is compressed from its equilibrium position to 0.50 meters , the force increases from 0 N to 10 N.
The average force = 10 ÷ 2 = 5 N

When the block is released, the spring exerts an average force of 5 N for 0.50 m.
Work done by spring = 5 * 0.50 = 10 N*m
This work increases the kinetic energy of the block!


The coefficient of friction between block and surface is 0.15.
Friction force = * mass * g

As the block moves 0.50 m, the work down by the friction force = * mass * g * d = - 0.15 * 0.500 * 9.8 * 0.50 = - 0.3675 N*m
The work done by friction is negative, because friction force decreases the kinetic energy of an object.

Net work = 10 + -0.3675= 9.6325 Joules

Kinetic energy = Net work
½ * 0.5* velocity^2 = 0.9.6325
Velocity = 6.207 m/s

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