A block with mass m =7 kg is hung from a vertical spring. When the mass hangs in

Question

A block with mass m =7 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.29 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? 236.7931 N/m

2) What is the oscillation frequency? .92566 Hz

3) After t = 0.39 s what is the speed of the block? _____m/s

4) What is the magnitude of the maximum acceleration of the block? 23.81 m/s2

5) At t = 0.39 s what is the magnitude of the net force on the block?_____ N

I just need 3 and 5 thank you!

Explanation / Answer

3) 0.5*m*V^2 = 0.5*K*A^2

A = 0.652 m

w= 2 pi f = 2x3.14x 0.92566= 5.81

y = A*sinwt = 0.652*sin(wt)

V= dy/dt = 0.652 x 5.81x cos(5.81 t)

At t= 0.39

Y= 0.652x sin(5.81 x 0.39) = 0.5 m

V= 0.652 x 5.81 cos(5.81 x 0.39) = 3.785 m/s

4. Correct

5. F= m y w2 = 7 x 0.5 x 5.812 = 118.146 N

Please rate the answer if found helpful..Goodluck

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