A spring with a spring constant of 8,000 N/m is used to accelerate a 1500 kg loaded elevator (including the passenger) upward at 3 m/s.

A) How far is the spring stretched?

B) If the passenger has a mass of 85 kg, what does a scale that he is standing on read?

 

Here we have given :

Spring constant of the spring : k = 8000 N/m

Mass of the elevator : M = 1500 kg

And, Acceleration of the elevator : a = 3 m/s² (Upward)

Part (A) Solution :

According to the Newton's Second law of motion : F_net = M a

Here, F_net = F_spring - M g

∴ F_spring - M g = M a

∴ F_spring = M a + M g

Since, F_spring = k x

∴ k x = M a + M g

∴ k x = M (a + g)

∴ (8000 N/m) x = (1500 kg)(3 m/s² + 9.81 m/s²)

∴ x = 2.4 m

Therefore, The spring will be stretched by : x = 2.4 m

Part (B) Solution :

Here, Mass of the passenger : m = 85 kg

Here, Reading in the scale will be equal to the normal force acting on the Passenger.

For the passenger : N - m g = m a

∴ N = m a + m g

∴ N = m (a + g)

∴ N = (85 kg)(3 m/s² + 9.81 m/s²)

∴ N = 1088.85 N

Therefore, The scale reading will be = 1088.85 N.

Please see the attached file for the complete solution