Two blocks, of masses M = 2.2 kg and 2 M , are connected to a spring of spring c
ID: 1323538 • Letter: T
Question
Two blocks, of masses M = 2.2 kg and 2M, are connected to a spring of spring constant k = 180 N/m that has one end fixed, as shown in the figure. The horizontal surface and the pulley are frictionless, and the pulley has negligible mass. The blocks are released from rest with the spring relaxed. (a) What is the combined kinetic energy of the two blocks when the hanging block has fallen 0.093 m? (b) What is the kinetic energy of the hanging block when it has fallen that 0.093 m? (c) What maximum distance does the hanging block fall before momentarily stopping?
Two blocks, of masses M = 2.2 kg and 2M, are connected to a spring of spring constant k = 180 N/m that has one end fixed, as shown in the figure. The horizontal surface and the pulley are frictionless, and the pulley has negligible mass. The blocks are released from rest with the spring relaxed. (a) What is the combined kinetic energy of the two blocks when the hanging block has fallen 0.093 m? (b) What is the kinetic energy of the hanging block when it has fallen that 0.093 m? (c) What maximum distance does the hanging block fall before momentarily stopping?Explanation / Answer
a)
Work done by gravity = 2MgS.
S is displacement of spring
kinetic energy gained by both masses
spring of spring constant k = 180
=2MgS-0.5ks2
=4.4*9.8*0.093-0.5*180*0.0932
=3.23J
b)
kinetic energy of the hanging block when it has fallen that 0.093 m
Since both the masses have the same velocity, their energy will be in the raio of their masses
KE1/KE2 =0.5*M*V2/0.5*2M*V2
KE1/KE2 =1/2
KE1+KE2 =3.23
3/2 *KE2=3.23
KE2=2.15J
c)
Smax is the maximum displacemet, both masses will loose their kinetic energies.
2MgSmax=0.5*K*Smax2
Smax=4Mg/k
=0.479m
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