Question

a mass of 0.75 kg is attached to a spring and placed on a horizontal surface. the spring has a spring constant of 180 N/m, and the spring is compressed 0.3 m past its natural length. if the mass is released from this compressed position, what is the speed of the mass as it passes the natural length of the spring?

Answer 1

Answer:

6.57 m/s

Explanation:

First use Hook's Law to determine the F the compressed spring acts on the mass. Hook's Law F=kx; F=force, k=stiffnes of spring (or spring constant), x=displacement

F=kx; F=180(.3) = 54 N

Next from Newton's second law find the acceleration of the mass.

Newton's .2nd law F=ma; a=F/m ; a=54/.75 = 72m/s²

Now use the kinematic equation for velocity (or speed)

v₂²= v₀² + 2a(x₂-x₀); v₂=final velocity; v₀=initial velocity; a=acceleration; x₂=final displacement; x₀=initial displacment.

v₀=0, since the mass is at rest before we release it

a=72 m/s² (from above)

x₀=0 as the start position already compressed

x₂=0.3m (this puts the spring back to it's natural length)

v₂²= 0 + 2(72)(0.3) = 43.2 m²/s²

v₂=$\sqrt{43.2)$ = 6.57 m/s