A block with mass m =6.8 kg is hung from a vertical spring. When the mass hangs

Question

A block with mass m =6.8 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.21 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.9 m/s. The block oscillates on the spring without friction.

1)

What is the spring constant of the spring?  

N/m

2)

What is the oscillation frequency?  

Hz

3)

After t = 0.28 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 = 0.28 s what is the magnitude of the net force on the block?  

N

6)

Where is the potential energy of the system the greatest?

At the highest point of the oscillation.

At the new equilibrium position of the oscillation.

At the lowest point of the oscillation.

Explanation / Answer

here,

mass of block = m = 6.8 kg
stretched disatnce of spring = X = 0.21 m
velocity given to block = 4.9 m/s

PART 1:
from newton second law we have :
Fnet = mg

where FNet = spring force , so
kx = mg

Sloving for spring constant, k
K = mg/x
K = 6.8 * 9.8 / 0.21
K = 317.33 N/m

PART 2:
Frequency of oscillationg spring is given as :

f = ( sqrt(K/m) ) / (2*pi)
f = ( sqrt(317.33 / 6.8) ) / (2*3.14)
f = 1.0877 Hz or 1.09 Hz

Part 3 :
velocity can be fuond out by using ,
v(t) = v * Cos(wt)

where
t = time
v = initial velocity = 4.9
w = angular speed = sqrt( g/x ) (in case of spring system)
w = sqrt(9.8/0.21)
w = 6.83 rad/s

at t = 0.28s
v(0.28) = 4.9*cos(6.83 * 0.28 )
v(0.28) = -1.641 m/s

This means that the mass is moving upward at -1.641 m/s.
Note that the argument of the Cos is in radians.

Part 4)
From conservation of erngy,

energy gained by block = potential energy stored in spring
0.5*m*v^2 = 0.5*k*del(x)^2

solving for max extension, del(x)
del(x) = sqrt(mv^2 / k)
del(x) = sqrt(6.8*4.9^2 / 317.33)
del(x) = 0.717

from newton second law we have :
Fnet = m*a

where FNet = spring force , so
k*del(x) = ma

solving for acceleration, a

a = k*del(x) / m
a = (317.33*0.717) / 6.8
a = 33.459 m/s^2 or 33.46 m/s^2

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