Physics question

Question

A spring with stiffness constant k = 72 N/m is compressed horizontally a total of 2.0 meters. A mass of 3 kg is placed in front of the spring and receives all of the spring's potential energy when it is released. If there is a friction force of 2 Newtons on the mass after it is released by the spring, how far in meters will the mass travel?  g=10

Answer 1

EPE of Spring = 1/2 (KX² )= 0.5 *72*2² = 144J

This energy is used against friction force

144 = Force *displacement

Displacement = 144/Force = 144 / 2 = 72 m

Answer 2

The easiest approach is a work energy approach:

E_Before + W_NC = E_after The mechanical energies are E_K,E_grav,pot.,E_spr and the only work is the negative work of friction: -F_fd. Because the system has no motion initially or after it stops, E_kin = 0. Since, there is no change in height, E_P = 0, and there is no spring potential after the mass is launched. The equation becomes

E_spr or 1/2kx² = F_fd (Granted, we could have gone directly to this equation in that we had spring energy that went into frictional losses.

d = kx²/(2F_f ) Everything is in MKS, so just plug in, although it's always a good idea to track the units.