A 950 g mass is hung on a “massless” vertical spring with spring constant k = 25

Question

A 950 g mass is hung on a “massless” vertical spring with spring constant k = 250 N/m.

a) How far does the mass stretch the spring when it is slowly lowered and

allowed to come to equilibrium? (2 pts)

b) The mass is now pulled downwards an additional 3.00 cm and let go so that it

starts to undergo simple harmonic motion about the equilibrium position.

Write down an expression for the total mechanical energy of the mass-spring

system at the point of release. (2 pts)

c) Write down an expression for the total mechanical energy of the mass-spring

system when the mass is 3.00 cm above the release point. (2 pts)

d) Using conservation of mechanical energy and your answers from parts b)

and c) determine the speed of the mass when it is 3.00 cm above the release

point. (2 pts)

e) Determine the velocity of the mass the first time it reaches a point that is 5.00

cm above the release point. (2 pts)

Explanation / Answer

a.) In the equilibrium position, the net upward force (tension in the spring) will be equal to the net downward force ( weight of the mass)

So, if the string id stretched by distance d, then

kd = mg

250 d = .950 x 9.81

d = 0.037278 m = 3.7278 cm

b.) at the lowest point the Kinetic and potentia energies are zero. Hence the total mechanical energy is the potential energy stored inside the spring which is

U = 0.5 k ( d + .03)2 = 0.5 x 250 x (.037278 + .03)2  = 0.5657911605 J

c.) at the position 3 cm above the release point, the stretch in the spring will be d and the potential energy of the spring will be U' = 0.5kd2   and the increase in the gravitational potential energy will be mgh and the Kinetic energy be KE = 0.5mv2   where v is the velocity at that point

So, the total mechanical energy will be the sum of the all potential and kinetic energies

Total mechanical energy = 0.5kd2 + mgh + 0.5mv2 = 0.5 x 250 x 0.0372782 + 0.95 x9.81 x0.03 + 0.5 x 0.95 x v2

d.) According to the law of conservation of energy the total mechanical energy should be the same at all the points

So, 0.5657911605 = 0.5kd2 + mgh + 0.5mv2 = 0.5 x 250 x 0.0372782 + 0.95 x9.81 x0.03 + 0.5 x 0.95 x v2

0.5657911605 = 0.5 x 250 x 0.0372782 + 0.95 x9.81 x0.03 + 0.5 x 0.95 x v2

0.5657911605 = 0.4532911605 + 0.475 v2

v = 0.4866642634 m/s = 48.66642634 cm /s

e.) at the point 5 cm above the release point, the stretch in the spring will be

d - 2cm = 3.7278 cm - 2cm = 1.7278 cm = 0.017278 m

Hence the potential energy stored inside the spring at this point will be

U'' = 0.5 x 250 x 0.0172782 = 0.0373161605 J

if v'' is the velocity at this point then the Kinetic energy will be 0.5 mv''2 = 0.5 x 0.95 x v''2  

and the increase in potential energy from the initital release point will be mgH = 0.95 x 9.81 x 0.05 = 0.465975 J

The total mechanical energy should be the same as in previous cases according to the law of conservation of energy

So,    0.5657911605 =  0.0373161605 + 0.5 x 0.95 x v''2   + 0.465975

v'' = 0.3627381251 m/s = 36.27381251 cm/s

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