1.Two crates of mass m1 and mass m2 are connected by a rope that passes over a p

Question

1.Two crates of mass m1 and mass m2 are connected by a rope that passes over a pulley (frictionless). Mass m1 is equal to 60 kg. What is the value of m2, if mass m1 accelerates downward at 4.5 m/s2?

2. A block on a frictionless horizontal surface is attached to a spring with a force constant (k) of 800 N/m. The block is pulled a distance of x to the right of its equilibrium position and then released from rest. If the kinetic energy of the block at the instant is passes through its equilibrium position is 4.5 J, what was the initial displacement from the equilibrium position?

Explanation / Answer

   1. We have the formula for the acceleration for the given situation as          a = g(m1 - m2) / (m1 + m2)          m2 = m1 (g - a)/ (a + g)          m2 = 60 (9.8 - 4.5) / (9.8 + 4.5)          m2 = 22.24 kg    2. The force constant of the spring, k = 800 N/m        The kinetic energy of the spring, ke = 4.5 J         We have, ke = (1/2) Kx^2                       (1/2) Kx^2 = ke                                    x = [2ke / K]^(1/2)                                    x = [2(4.5) / 800]^(1/2)                                    x = 0.11 m    The initial displacement from the equilibrium position, x = 0.11 m          m2 = 60 (9.8 - 4.5) / (9.8 + 4.5)          m2 = 22.24 kg    2. The force constant of the spring, k = 800 N/m        The kinetic energy of the spring, ke = 4.5 J         We have, ke = (1/2) Kx^2                       (1/2) Kx^2 = ke                                    x = [2ke / K]^(1/2)                                    x = [2(4.5) / 800]^(1/2)                                    x = 0.11 m    The initial displacement from the equilibrium position, x = 0.11 m          m2 = 22.24 kg    2. The force constant of the spring, k = 800 N/m        The kinetic energy of the spring, ke = 4.5 J         We have, ke = (1/2) Kx^2                       (1/2) Kx^2 = ke                                    x = [2ke / K]^(1/2)                                    x = [2(4.5) / 800]^(1/2)                                    x = 0.11 m    The initial displacement from the equilibrium position, x = 0.11 m                                    x = [2ke / K]^(1/2)                                    x = [2(4.5) / 800]^(1/2)                                    x = 0.11 m    The initial displacement from the equilibrium position, x = 0.11 m                                    x = 0.11 m    The initial displacement from the equilibrium position, x = 0.11 m

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