A block with mass m =7.4 kg is hung from a vertical spring. When the mass hangs

Question

A block with mass m =7.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.23 m. While at this equilibrium position, the mass is then given an initial push downward at v = 3.9 m/s. The block oscillates on the spring without friction.

1)What is the spring constant of the spring? --->> the answer (315.30)

2)What is the oscillation frequency? --->>> the answer ( 1.03)

I NEED HELP WITH THE NEXT QUESTIONS.

3) After t = 0.42 s what is the speed of the block?

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

5) At t = 0.42 s what is the magnitude of the net force on the block?

6) Where is the potential energy of the system the greatest? YOU CAN CHOOSE MULTIABLE ANSWERS

A) At the highest point of the oscillation.

B) At the new equilibrium position of the oscillation.

C) At the lowest point of the oscillation.

Explanation / Answer

mass m= 7.4 kg
x =0.23 m
1)
315.30 N/m
---------------
2)
freq f = 1.03 Hz
----------------
3)
ang freq w = 2pi*f = 6.28*1.03 = 6.4684 rad/s
from given data V= wA
A = 3.9/6.4684 = 0.6029 m
from
x = A sinwt
v = wA cos(wt)
at t =0.42 sec
v = (6.4684)0.6029*cos(6.4684*0.42) = -3.55 m/s
-------------------------
4)
max acceleration
a =-w^2*A = -25.22 m/s^2
------------------------
5)
at t=0.42 sec
position x = 0.248 m
acceleration a = -w^2*x = -10.4 m/s^2
force F = ma = 7.4*(-10.4) = 76.9 N
-------------------------
6)
potential energy =0.5*kA^2 = 0.5*315.30*0.6029^2 = 57.30 J

A) At the highest point of the oscillation.

C) At the lowest point of the oscillation.
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ALL THE BEST

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