Please show work. A spring with force constant A; = 400-N/m is contracted 20 cm
ID: 1425676 • Letter: P
Question
Please show work. A spring with force constant A; = 400-N/m is contracted 20 cm from equilibrium. It is released from rest at t = 0 (as in the figure) to push an M = 1.0-kg mass placed against the spring, till M reaches the spring's equilibrium at the edge of the table, where M is launched horizontally into the air 1.25 m above the floor. How much potential energy is stored in the spring (at t = 0)? How much kinetic energy does the mass achieve at launch? What is the speed of M at release? What is the magnitude of the momentum of M at launch How long does it take M to reach the floor? How far from the foot of the table is M when it hits the floor? What is the kinetic energy of M just as it reaches the floor?Explanation / Answer
K = 400 N/m
x = 20 cm = 0.2 m
(13)
Potential Energy stored in the spring, U = 1/2*kx^2
U = 1/2*400*0.2^2
U = 8 J
(14)
Due to Energy Conservation.
Kinetic Energy = Same as above = 8J
(15)
K.E = 8 J
1/2 * mv^2 = 8
v^2 = 8*2 / 1 m/s
v = 4 m/s
Speed at release , v = 4 m/s
(16)
Momentum = M*v
P = 1.0 * 4 Kgm/s
P = 4.0 Kg m/s