A block with mass m =6.2 kg is hung from a vertical spring. When the mass hangs
ID: 2108447 • Letter: A
Question
A block with mass m =6.2 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.22 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.7 m/s. The block oscillates on the spring without friction.
*******1) What is the spring constant of the spring? ___________ N/m.....
******2)What is the oscillation frequency? ____________ Hz......
******3) After t = 0.33 s what is the speed of the block? __________m/s....
******4)What is the magnitude of the maximum acceleration of the block? ________m/s^2....
******5)At t = 0.33 s what is the magnitude of the net force on the block? __________N.... ****************************************************************************************************************************************
A block with mass m =6.2 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.22 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.7 m/s. The block oscillates on the spring without friction.
*******1) What is the spring constant of the spring? ___________ N/m....
******2)What is the oscillation frequency? ____________ Hz......
******3) After t = 0.33 s what is the speed of the block? __________m/s....
******4)What is the magnitude of the maximum acceleration of the block? ________m/s^2....
******5)At t = 0.33 s what is the magnitude of the net force on the block? __________N..
Explanation / Answer
1) mg=K*x
==> K=276 .18 N/m
2) occilating frequency=sqrt(K/m)/2*pi=1.0627 Hz
3) v=4.7 cos(2*pi*1.0627*t)
at t=0.33
v=4.7 m/s
it is too difficult to decode this Question...
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