Question

A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. When the spring is 4.0 cm longer than its equilibrium length, the speed of the block is 0.50 m/s. The greatest speed of the block is:

Answer 1

The greatest speed is 0.711 m/s

Step-by-step explanation:

The total energy is only kinetic energy (no elongation in the spring), thus, the greatest speed of the block is equal:

`(1)/(2) mv_(i) ^(2) +(1)/(2) ks^(2) =(1)/(2) mv_(f) ^(2)`

Where

m = 0.5 kg

vi = 0.5 m/s

k = 80 N/m

s = 4 cm = 0.04 m

Replacing:

`(1)/(2) *0.5*0.5^(2) +(1)/(2) *80*0.04^(2) =(1)/(2) *0.5*v_(f) ^(2) \\0.0625+0.064=0.25v_(f) ^(2) \\v_(f) =\sqrt{(0.0625+0.064)/(0.25) } =0.711m/s`

Answer 2

Answer: 0.711 m/s

Step-by-step explanation:

Given

Mass of the block, m = 0.5 kg

Force constant of the spring, k = 80 N/m

Extension of the spring, x = 4 cm = 0.04 m

Speed of the block, v = 0.5 m/s

Now, we know that the energy in the spring is = 1/2kx² and that the kinetic energy is 1/2mv²

From the question, our initial time for our initial energy has some kinetic energy and some spring energy.

This, at the time of it's greatest speed, at the final time or final energy, it will be at equilibrium, x=0 and will have all kinetic energy. If we use law of conservation of energy, we know that Initial Energy is equal to final energy

Thus, the spring energy + the kinetic energy at 4 cm = the kinetic energy at 0 cm.

Therefore,

1/2kx² + 1/2mv² = 1/2mv(f)², where v(f) is the greatest speed of the block

1/2 * 80 * 0.04² + 1/2 * 0.5 * 0.5² = 1/2 * 0.5 * v(f)²

1/2 * 0.128 + 1/2 * 0.125 = 1/2 * 0.5 v(f)²

0.064 + 0.0625 = 0.25 v(f)²

0.1265 = 0.25 v(f)²

v(f)² = 0.506

v(f) = √0.506

v(f) = 0.711 m/s

Thus, the greatest speed of the block is 0.711 m/s