Two blocks, of masses M = 2.3 kg and 2M, are connected to a spring of spring con

Question

Two blocks, of masses M = 2.3 kg and 2M, are connected to a spring of spring constant k = 200 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 0.092 m? (b) What is the kinetic energy of the hanging block when it has fallen that 0.092 m? (c) What maximum distance does the hanging block fall before momentarily stopping? Number Units Number Units Number Units

Explanation / Answer

given that

M = 2.3 kg

k = 200 N/m

(a)

x = 0.092 m

potential energy due to gravity will work at 2*M block

h will be equal to displacement of spring = x = 0.092 m

PEg = 2*M*g*x

using energy conservation

PEg = KE of block M + kE of block 2*M + PE of spring

PEg = M*v^2/2 + 2*M*v^2/2 + k*x^2/2

M*v^2/2 + 2*M*v^2/2 = 2*M*g*x - k*x^2/2

KE of both the blocks = 2*2.3*9.8*0.092 - 200*(0.092)^2/2

KE of both the blocks = 3.30 J

(b)

both the blocks have the same velocity.so their kinetic energy will be in the raio of their masses

2*M*v^2/2 / M*v^2/2 = 2/1

kinetic energy of M is = (1/3)* 3.30 = 1.10 J

kinetic energy of 2*M = (2/3)*3.30 = 2.20 J

(c)

there is no change in total KE and KE is converted into PE

so PEg = PE of spring

2*M*g*xmax = k*x^2 / 2

where xmax = maximum displacement

xmax = 4*M*g/k = 4*2.3*9.8/200

xmax = 0.45 m

answer

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