Aglider of mass 0.170 kg moves on a horizontal frictionless air track. it is permanently attached to one end of a massless horizontal spring, which has a force constant of 13.0 n/m both for extension and for compression. the other end of the spring is fixed. the glider is moved to compress the spring by 0.180 m and then released from rest.(a) calculate the speed of the glider at the point where it has moved 0.180 m from its starting point, so that the spring is momentarily exerting no force. m/s(b) calculate the speed of the glider at the point where it has moved 0.250 m from its starting point. m/s

a.  1.594 m/s = v

b. 1.274 m/s = v

Explanation:

A) First calculate the potential energy stored in the spring when it is compressed by 0.180 m...

U = 1/2 kx²

Where U is potential energy (in joules), k is the spring constant (in newtons per meter) and x is the compression (in meters)

U = 1/2(13.0 N/m)(0.180 m)² = 0.2106 J

So when the spring passes through the rest position, all of its potential energy will have been converted into kinetic energy.  K = 1/2 mv².

 0.2106 J  = 1/2(0.170 kg kg)v²

0.2106 J  = (0.0850 kg)v²

2.808m²/s² = v²

1.594 m/s = v

(B)  When the spring is 0.250 m from its starting point, it is 0.250 m - 0.180 m = 0.070 m past the equilibrium point.  The spring has begun to remove kinetic energy from the glider and convert it back into potential.  The potential energy stored in the spring is:

U = 1/2 kx² = 1/2(13.0 N/m)(0.070 m)² = 0.031J

Which means the glider now has only 0.2106 J  - 0.031J = 0.1796 J of kinetic energy remaining.

K = 1/2 mv²

0.1796 J = 1/2(0.170 kg)v²

0.138 J = (0.0850 kg)v²

1.623 m²/s² = v²

1.274 m/s = v

explanation:

as the height from which the ruler was dropped increased, the speed of the ruler at ryan's hand increased.