Question

A block with mass m =7.5 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.25 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.

A block with mass m-7.5 ka hung from a vertical spring. Whon the ms hang in equilibrlum, the spring stretches x 0.25m. while at thk equErium pacition, the mass is then given an initial puch downwand at v-4.1 mi's. The block cluanes on the spring without friction 1) Whar is the spring constant of the spring Your ubmbsiora: 943 Computed value: 294 .3 ubitbed: Salurday, Dcember 2 al 8:52 Pm Whar is the acillacon frequency? 31 After t-0.3 s what k thespood d the block? Att-0.3 s what &the magnitude of the net force on the block 6) Whore i the potencial oneay f the system the greates D At the highest paint of the ascillation. DAt the new equilibrium position of the cscillation At the lowest paint of the ecillation.

Answer 1

Given, m = 7.5 kg, x = 0.25 m , vo = 4.1 m/s

a) mg = kx

k = mg/x = 7.5*9.8/0.25

k = 294 N/m

b) f = (1/2π)*√(m/k)

f = (1/2π)*√(7.5/294)

f = 0.025 Hz

c) At t = 0.3 s ,

Using , v = -vo*cos(wt)

- sign indicates it is going down

Since , w = √(k/m) = √(294/7.5) = 6.261 rad/s

So,

v = - 4.1*cos(6.261*0.3)

v = 1.241 m/s

d) Maximum acceleration (Amax) = vo*w

Amax = 4.1*6.261

Amax = 25.67 m/s^2

E)Now , amplitude (A) = vo/w

A = 4.1/6.261 = 0.655 m

y = -Asin(wt)

y = -0.655*sin(6.261*0.3)

y = -0.624 m

Now,

Fnet = m*w^2*|y|

Fnet = 7.5*6.261^2*0.624

Fnet = 183.457 N

F) At the highest point of oscillation , the potential energy is greatest