2. A massless spring with a 1.5 kg mass attached is suspended vertically an addi

Question

2. A massless spring with a 1.5 kg mass attached is suspended vertically an additional force
of – 10 N is applied to the system. The overall displacement of the spring from its
unstrained equilibrium position is 50 cm:

a. Determine the spring constant

b. Determine the equilibrium position of the spring when only the weight is
attached. I other words the position of the spring when the weight is just hanging
on the spring and not moving

c. Determine the velocity of the 1.5 kg mass when it passes through the equilibrium
position of the spring with the weight attached , when it is traveling upward.

Explanation / Answer

2.

x = stretch in the spring = 50 cm = 0.50 m

k = spring constant

F = spring force = weight of mass attached + additional force

k x = 1.5 x 9.8 + 10

k (0.50) = 1.5 x 9.8 + 10

k = 49.4 N/m

b)

x' = stretch in spring due to weight only

kx' = mg

49.4 x' = (1.5) (9.8)

x' = 0.298 m

c)

KE = spring potential energy

(0.5) m v2 = (0.5) k x'2

1.5 v2 = (49.4) (0.298)2

v = 1.71 m/s

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