A massless spring with a 2.2 kg mass attached is suspended vvertically. An addit

Question

A massless spring with a 2.2 kg mass attached is suspended vvertically. An additional force -25 N is applied to the system. The overall displacement of the spring from its unstrained equilibrium position is 110cm.

a.) determine the spring constant

b.) Determine the equilibrium position of the spring when only the weight is attatched.

c.) What is the net force experience by the mass when it is released?

d.) What is the force applied by the spring of the mass?

e.) What is the acceleration experience by the mass when it is released?

f.) What distance from the unstrained equilibrium point will the acceleration be 0?

g.) What distance from the strained equilibriumpoint will the acceleration be 0?

h.) What distance from the unstrained equilibrium point will the mass reach its max velocity?

i.) What distance from the strained equilibrium point will the mass reach its max velocty?

j.) What is the frequency of oscillation of the ssytem in Hz, if the mass were released?

Explanation / Answer

Here,

mass , m = 2.2 Kg

F = -25 N

x = 110 m

a) let the spring constant is k

as k * 1.1 = 25 + 2.2 * 9.8

k = 42.33 N/m

the spring constant is 42.33 N/m

b)

let the equilibrium position of weight is x

W = m * g = k * x

2.2 * 9.8 = 42.33 * x

x = 0.51 m

the equilibrium position of the spring when only the weight is attatched is 0.51 m

c)

for the net force acting on the mass

net force acting on the mass = k * x - m * g

net force acting on the mass = 42.33 * 1.1 - 2.2 * 9.8

net force acting on the mass = 25 N

d)

for the force applied by the spring of the mass

force applied by the spring of the mass = k * x

force applied by the spring of the mass = 42.33 * 1.1 N

force applied by the spring of the mass = 46.6 N

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