Question

A 2.15 kg block on a horizontal floor is attached to a horizontal spring that

is initially compressed 0.0320 m. The spring has force constant
900 N/m. The coefficient of kinetic friction between the floor and the
block is 0.400. The block and spring are released from rest and the block
slides along the floor.

What is the speed of the block when it has moved a distance of 0.0180 m from its initial position? (At this point the spring is compressed .0140 m)

Answer 1

The speed of the block when it has moved a distance of 0.0180 m from its initial position is 0.4531 m/s.

What is the work done by the spring?

When the block has moved a distance of 0.0180 m from its initial position, the spring is compressed 0.0140 m. The work done by the spring is:

work done by spring = 0.5 * spring constant * (initial_compression 2 - final_compression2)

work_done_by_spring = 0.5 * 900 * (0.0320**2 - 0.0140**2)

work_done_by_spring = 1.728 J

The work done by friction is:

normal_force = mass * 9.81 # N

friction_force = coefficient_of_friction * normal_force

friction_force = 0.400 * 2.15 * 9.81

friction_force = 8.436 N

distance_moved = 0.0180 m

work_done_by_friction = friction_force * distance_moved

work_done_by_friction = 0.152 J

The total work done is:

total_work_done = work_done_by_spring - work_done_by_friction

total_work_done = 1.728 J - 0.152 J

total_work_done = 1.576 J

The kinetic energy of the block is:

kinetic_energy = total_work_done

kinetic_energy = 1.576 J

The speed of the block is:

speed =
`\sqrt{}`(2 * kinetic_energy / mass)

speed =
`\sqrt{}`(2 * 1.576 J / 2.15 kg)

speed = 0.4531 m/s

Therefore, the speed of the block when it has moved a distance of 0.0180 m from its initial position is 0.4531 m/s.