A block of mass m = kg rests on a frictionless floor. It is attached to a spring

Question

A block of mass m = kg rests on a frictionless floor. It is attached to a spring with a relaxed length L = m. The spring has spring constant k = N/m and is relaxed when hanging in the vertical position. The block is pulled d = m to one side. In this problem, the block is always constrained to move on the floor (i.e. it never leaves the floor).

By what amount is the spring extended?
What is the potential energy stored in the spring?
The block is released but is constrained to move horizontally on the frictionless floor. What is the maximum speed it attains?

Let's change the problem a bit. When the spring is vertical (hence, unstretched), the block is given an initial speed equal to 1.4 times the speed found in part

How far from the initial point does the block go along the floor before stopping?

How far from the initial point does the block go along the floor before stopping?

Explanation / Answer

Spring extension x = (d^2 + L^2) - L

Potential energy stored in spring = 0.5*kx^2 = 0.5*k*((d^2 + L^2) - L)^2

For max. speed, all potential energy should be converted into kinetic energy. Hence,

0.5*mv^2 = 0.5*kx^2

v = x*(k/m)^0.5

v = ((d^2 + L^2) - L)*(k/m)^0.5

New velocity V = 1.4*v = 1.4*((d^2 + L^2) - L)*(k/m)^0.5

For max. distance moved, all KE should be converted into PE.

0.5*mV^2 = 0.5*kX^2

New extension in spring X = V*(m/k)^0.5

X = 1.4*((d^2 + L^2) - L)

New Extended spring length = L+X = 1.4*(d^2 + L^2) - 0.4*L

Horizontal distance moved = [1.4*(d^2 + L^2) - 0.4*L]^2 - L^2

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