A spring hangs vertically, A 250 g mass is attached to the spring and allowed to

Question

A spring hangs vertically, A 250 g mass is attached to the spring and allowed to come to The spring stretches 8 cm as the mass comes to rest. What is the spring constant of the a) 2.5 N/m b) 3.1 N/m c) 31 N/m d) 122 N/m e) 250 N/m The mass is moved up to a Position 3 cm above its equilibrium position on the spring and released. At the moment that the mass is released, what is the magnitude of the net force on mass? a) 0 b) 0.92 N c) 2.5 N d) 3.4 N e) 9.8 N What is the direction of the net force in #1.2? up down zero (circle one) What is the angular frequency of oscillations of the mass on the spring. a) 1.8/s b) 5.5/s c) 11/s d) 18/s e) 122/s From the time that the mass is released, how long is it until the next time that the mass has zero velocity? a) 142 ms b) 174 ms c) 284 ms d) 449 ms e) 658 ms What is the magnitude of the acceleration of the mass as it passes through equilibrium? a) 0 b) 0.33 m/s^2 c) 3.7 m/s^2 d) 9.8 m/s^2 e) 56 m/s^2

Explanation / Answer

3. Fnet = k x - m g = 0

k (0.08) = 0.250 x 9.81

k = 30.66 N/m

Ans(C)

4. F = (30.66) (0.03) = 0.92 N

Ans(b)

5. down

6. w = sqrt(k / m)

= sqrt(30.66 / 0.250)

= 11.1 rad/s

Ans(C)

7. it will after t = T/2

where T = time period = 2 pi / w = 0.566 sec Or 566 msec

t = 566 / 2 = 283 ms

Ans(c)

8. at equlibrium, Fnet = 0

and a = F / m

a = 0

Ans(a)

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