Posted by Student B on .
A light spring having a force constant of 125 N/m is used to pull a 9.50 kg sled on a horizontal ice rink. The coefficient of kinetic friction between the sled and the ice is 0.200. The sled has an acceleration of 2.00 m/s^2.
By how much does the spring stretch if it pulls on the sled horizontally?
By how much does the spring stretch if it pulls on the sled at 30.0 above the horizontal?
Horizontal case first:
The spring must provide a force F that makes the net force (spring minus friction) equal to M*a.
F - M*g*u = M*a
(u is the kinetic friction coefficient; a is the acceleration)
F = M*(a + u*g)
Once you have F, the stretch is F/k,
where k is the spring contant.
With the spring at a 30 degree angle, only the horizontal component does the accelerating, but the vertical component will reduce the friction somewhat.
F cos30 - (M*g-Fsin30)*u = M*a
Solve for F etc.