A force of magnitude 41.0 N stretches a vertical spring a distance 0.240 m. A) W

Question

A force of magnitude 41.0 N stretches a vertical spring a distance 0.240 m.

A) What mass must be suspended from the spring so that the system will oscillate with a period of 1.04 s ?

B) If the amplitude of the motion is 5.10×102 m and the period is that specified in part (a), where is the object at a time 0.340s after it has passed the equilibrium position, moving downward? (Take the upward direction positive.)

C) What force (magnitude) does the spring exert on the object when it is a distance 3.20×102 m below the equilibrium position, moving upward?

Take the free fall acceleration to be 9.80 m/s^2.

D) Find the direction of this force. Upward or downward?

Explanation / Answer

a)

Here , as Force = k * x

41 = 0.240 * k

k = 170.8 N/m

let the mass is m

T = 2pi * sqrt(m/k)

1.04 = 6.282 * sqrt(m/170.8)

m = 4.68 Kg

the mass suspended is 4.68 Kg

B)

y = 0.051 * sin(2pi * 0.34/1.04)

y = 0.045 m

the block is 0.045 m below the equilibrum position

c)

let the force is F

F = 170.8 * 3.2 *10^-2

F = 5.56 N

d)

the direction of the force is downwards

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