A 0.5 kg mass is connected to the lower end of an ideal vertical coil spring sus

Question

A 0.5 kg mass is connected to the lower end of an ideal vertical coil spring suspended from a retort stand. The mass is held at rest with the spring completed unstretched and then allowed to fall.

a) If the spring constant is 20 N/m, determine the maximum distance the mass drops before coming momentarily to rest and starting up again. Ignore frictional losses.

b) In the problem above, the mass is now held initially at rest at a position where the spring is stretched 20 cm. Determine the maximum distance the mass will fall. (0.1 m)

c) Determine how fast the mass was travelling on the way down when the spring was stretched 30 cm.

Explanation / Answer

We can use the conservation of energy to find the distances
We know that the total energy must be conserved for an isolated system ,consider the total energies at two instants
,We can take the gravitational potential energy to be zero at the unstretched position. The mass is at rest so the KE is also zero
At the unstretched position the total energy is zero
Consider the second instant when the mass is momentarily at rest after falling a distance x
The gravitational potential energy is at the position is mg(-x)
The elastic potential energy of the spring is (1/2) k x2
The total energy at x is (1/2)kx2 - mgx
equating the total energies at two positions
   (1/2)kx2 - mgx = 0
(1/2)kx   = mg
x = 2 m g / k = 2 x 0.5 x 9.8 / 20
x = 0.49 m

b) Again we can use the conservation of energy
Consider the instant when the spring is stretched 20 cm
We can take the gravitational potential energy to be zero .
The elastic potential energy at the position is (1/2) k (0.2)2
The KE is zero because the mass is at rest
E = (1/2) k x 0.22
At the second instant when the mass is momentarily at rest after falling a distance x' from the stretched position
The gravitational potential energy is mg(-x')
The elastic potential energy is 1/2 k (x' + 0.2)2
E = 1/2 k (x' + 0.2)2 - mgx'
Equating the total energies at the two positions
(1/2) k x 0.22   =   1/2 k (x' + 0.2)2 - mgx'
x'2 + 0.4 x' - 2 x' mg / k = 0
This is a quadratic equation of x'
Solving this we get
x' = 0 ,or x' = 0.09 m ~ 0.1 m

c) We can again use the conservation of energy
The first instant is when the mass is at the unstretched position
The gravitational ,elastic and the kinetic energies are zero there
The second instant is when the mass is traveling with a velocity v ,at a distance x = 0.3m
The elastic potential energy is (1/2) kx2
The gravitational potential energy is mg (-x)
The KE at the instant is (1/2) m v2
Equating the total energies at both the times
(1/2) kx2  + (1/2) mv2 - mgx = 0
0.5 x 20 x (0.3)2  + 0.5 x 0.5 x v2   = 0.5 x 9.8 x 0.3
Solving for v
v = 1.5099 m /s

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