A single mass (m 1 = 3.2 kg) hangs from a spring in a motionless elevator. The s

Question

A single mass (m1 = 3.2 kg) hangs from a spring in a motionless elevator. The spring constant is k = 347 N/m.

1) What is the distance the spring is stretched from its unstretched length? cm

Now, three masses (m1 = 3.2 kg, m2 = 9.6 kg and m3 = 6.4) hang from three identical springs in a motionless elevator. The springs all have the same spring constant given above.

3) What is the distance the middle spring is stretched from its equilibrium length? cm

Now the elevator is moving downward with a velocity of v = -3.2 m/s but accelerating upward at an acceleration of a = 4 m/s2. (Note: an upward acceleration when the elevator is moving down means the elevator is slowing down.)

5)What is the distance the lower spring is extended from its unstretched length? cm

Finally, the elevator is moving downward with a velocity of v = -3.7 m/s but accelerating downward at an acceleration of a = -2.4 m/s2.

7)What is the magnitude of the net force on the middle mass? N

Explanation / Answer

Using the Hook’s law F = -k*x and considering the forces due to gravity and the acceleration of the elevator on the masses we can find the change in length of the spring.

Part 1) The mass m1 = 3.2 kg is attached to the spring of the spring constant k = 347N/m, the spring is attached to ceiling of the motionless elevator. The weight of mass m1; W = m1*g acts on the spring causing it to stretch.

W = 3.2kg*(-9.8m/s2)

W = - 31.36 N                     (The negative sign indicates the downward direction of weight)

Thus the net force on the spring is F = W = -31.36N

F = -k*x

-31.36N = -(347N/m)* x

x = 0.09037 m

x = 9.037 cm

The spring is stretched by 9.037 cm from its un-stretched length.

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Part 3) The three masses (m1 = 3.2 kg, m2 = 9.6 kg and m3 = 6.4 kg) hang from three identical springs in a motionless elevator. The masses m2 hangs from middle spring and the mass m3 hangs below mass m2.

Thus middle spring supports the masses m2 and m3. The weight of m2 and m3 acts on the middle spring causing it to stretch.

W23 = (m2+m3)*g

W23 = (9.6 kg + 6.4 kg)*(-9.8m/s2)

W23 = - 156.80 N

Now F = W23 = -k*x

- 156.80 N = -(347 N/m)* x

x = 0.45187 m

x = 45.187 cm

Thus the middle spring is stretched from its equilibrium position by 45.187 cm.

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Part 5) The lower spring is attached only to the m3 = 6.4kg alone. The weight due to m3 causes the spring to stretch.

In addition to this, the elevator is moving downwards and slowing down, i.e. the elevator has an upward acceleration. This causes the mass m3 to accelerate in downward direction (since it is not rigidly fixed with elevator) Thus the net force on the spring is given by

F = (g + a)*m3

F = (-9.8m/s2 – 4 m/s2)*6.4kg

F = -88.32 N

Thus F = -k*x gives

-88.32 N = - (347 N/m)* x

x = 0.2545 m

x = 25.45 cm

Thus the lower spring is extended by 25.45 cm from its un stretched length.

Part 7) The mass m3 hangs below m2 with the help of a spring.

Now the elevator is moving downwards and is accelerating in downward direction.

The gravitational force acts on the masses in downward direction. But the elevator is accelerating in downward direction (a = -2.4m/s2) causing the mass m2 and m3 to accelerate in upward direction a’ = 2.4m/s2. The net force on the middle mass is the sum of the weights of m2 and m3 minus the upward force due to elevators acceleration;

F = (m2+m3)*g – (m2+m3)*a’

F = (9.6 kg+6.4kg)*(-9.8 m/s2) – (9.6 kg+6.4kg)*(2.4 m/s2)

F = - 118.4 N

Thus the magnitude net force on the middle mass is 118.4 N

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