Question

2. Consider the following physical situation: A spring that obeys Hooke's Law and has a known/given spring constant k has been compressed to half of its equilibrium length. It's anchored at one end while the other end pushes (but is not attached to) a block of mass m in the horizontal direction. The block is initially held in place. Once released, the block accelerates to the right and achieves a final speed of ve at the point when it leaves contact with the spring. We'll call the horizontal position of the block "x" and define x 0m as the block's starting position (so the spatial origin is not the position of the block when the spring it at equilibrium, it's the block's initial position which occurs when the spring is compressed to half of its equilibrium length). We'll also call the equilibrium length of this spring 2L so its starting (compressed) length is L (a) What is the magnitude of the spring force on the block (clearly it is acting to the right) when the block is at positionx (for 0sx sL)? (b) Calculate the total work done on the block by this spring as it decompresses. Note: The (spring) force points in the same direction as the displacement of the block. c) Use the work-energy theorem to calculate the final speed of the block (v:). (d) Use conservation of energy to calculate ve and compare this value to your answer from (c).

Answer 1

a) The spring is initially compressed 'L' meters, when the block is x meters from the position, the compression of the spring is, (L-x) meters, hence according to Hooke's law, the force is, the product of spring constant and the compression

F=  k*(L-x)

b) The work done by the spring is the change in its potential energy,

W = U_f - U_i

potential energy is given by,

U = 0.5kX²

k is the spring constant and X is the compression

initially when it is compressed, its potential energy, U_i = 0.5kL²

finally when it is completely decompressed, (compression=0) , its potential energy, U_f = 0.5k*0² = 0

hence work done by the spring, W = 0 - 0.5kL²

W = -0.5kL²

-ve only represents that the potential energy of the block decreases

c) Work done is equal to the change in kinetic energy,

W= 0.5mv_f²

0.5kL² = 0.5mv_f²

v_f = sqrt(kL²/m)

d) The potential energy of the spring is converted to the kinetic energy of the kinetic energy of the block

U = K

0.5kL² = 0.5mv_f²

v_f = sqrt(kL²/m)