A block with mass m =7.5 kg is hung from a vertical spring. When the mass hangs

Question

A block with mass m =7.5 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.25 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.1 m/s. The block oscillates on the spring without friction.

1)

What is the spring constant of the spring?

294.3 N/m  

2)

What is the oscillation frequency?

.997 Hz  

3)

After t = 0.3 s what is the speed of the block?

m/s  

4)

What is the magnitude of the maximum acceleration of the block?

m/s2  

5)

At t = 0.3 s what is the magnitude of the net force on the block?

N

Explanation / Answer

1) m g = k x

k = m g/x = 7.5*9.81/0.25= 294.3

2) f = 1/(2*pi)*sqrt(k/m) =1/(2*pi)*sqrt(294.3/7.5)= 0.997

3)

v = vmax cos( 2 pi f t) =4.1*cos(2*pi*0.997*0.3)= -1.24 m/s

4) 1/2 k A^2 = 1/2 mv^2

A = sqrt( 7.5*4.1^2/294.3)= 0.655 m

max a = k A/m = 294.3*0.655/7.5= 25.7 m/s^2

5) a = 4.1*2*pi*0.997*sin(2*pi*0.997*0.3)= 24.47 m/s^2

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