A block with mass m =6.7 kg is hung from a vertical spring. When the mass hangs

Question

A block with mass m =6.7 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.27 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.5 m/s. The block oscillates on the spring without friction.

1)What is the spring constant of the spring? 243.4 N/m

2)What is the oscillation frequency? 0.95811275851 Hz

3)After t = 0.46 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 = 0.46 s what is the magnitude of the net force on the block? -----------------------N

Explanation / Answer

1)

spring constant K = mg/x = 6.7*9.8/0.27 = 243.4 N/m

(2)

angular frequency w = sqrt(k/m) = sqrt(243.4/6.7) = 6.03 rad/s

f = w/(2pi) = 6.03/(2*pi) = 0.96 Hz

(3)

equation of motion

y = A*sin(wt + pi)

speed v = dy/dt = A*w*cos(wt + pi)

acceleration a = dv/dt = -A*w^2*sin(wt + pi) = -w^2*y

given speed at equilibrium = Aw = 4.5 m/s

at t = 0.46

v = 4.5*cos(6.03*0.46+pi)

v = 4.2 m/s

===============

(4)

amax = w^2*A = w*(w*A) = 6.03*4.5 = 27.135 m/s^2

==============

Fnet = m*a

Fnet = m*w^2*A*sin(wt+pi)

Fnet = 6.7*27.135*sin((6.03*0.46)+pi)

Fnet = 65.37 N <<<<<<=========ANSWER

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