One end of a massless coil spring is suspended from a rigid ceiling. It is found that when a 3.3 kg mass is suspended from it the spring stretches 105 mm when the mass comes to rest.
(a) What is the spring constant for the spring?
(b) If the mass is then pulled down an additional 165 mm (by hand) and released from rest at that point, what will be the speed of the mass when it passes through the spring-mass rest position (this is where the spring was stretched 100.mm by the stationary mass).
(a) k = Weight/Deflection
= M g/(0.105 m) = 308 N/m
(b) V at that time will be whatever makes the kinetic energy equal to the spring potential energy at maximum displacement from the equilibrium position. Refer spring potential energy to the equilbrium position.
(1/2)k X^2 = (1/2) M V^2
X = 0.165 m
M = 3.3 kg
Solve for V
To find the spring constant for the given mass-spring system and determine the speed of the mass when it passes through the spring-mass rest position, we can use Hooke's Law and the principle of conservation of mechanical energy.
(a) First, we need to find the spring constant. Hooke's Law states that the force exerted by a spring is directly proportional to the amount it is stretched or compressed. Mathematically, it can be expressed as:
F = -k * x
where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the spring stretches by 105 mm (or 0.105 m) when a 3.3 kg mass is suspended from it. The weight of the mass (mg) provides the force needed to stretch the spring, so we have:
F = mg
Substituting the known values, we get:
mg = -k * x
k = -mg / x
k = -(3.3 kg * 9.8 m/s^2) / 0.105 m
Calculating the expression gives us the spring constant:
k ≈ -305.71 N/m
Note that the negative sign indicates that the force exerted by the spring is in the opposite direction to the displacement.
(b) To find the speed of the mass when it passes through the spring-mass rest position, we can use the conservation of mechanical energy. At the initial point (when the mass is pulled down an additional 165 mm), the potential energy is converted into kinetic energy as it falls. At the spring-mass rest position, all the potential energy is converted into kinetic energy.
The potential energy at the initial position is given by:
PE1 = m * g * h1
where h1 is the initial displacement.
The potential energy at the spring-mass rest position is given by:
PE2 = 0.5 * k * x^2
where x is the displacement from the equilibrium position (100 mm or 0.1 m).
Since the total mechanical energy remains constant, we have:
PE1 = PE2 + KE
where KE is the kinetic energy at the spring-mass rest position.
Rearranging the equation, we get:
KE = PE1 - PE2
KE = m * g * h1 - 0.5 * k * x^2
Substituting the known values:
- m = 3.3 kg
- g = 9.8 m/s^2
- h1 = 0.165 m (initial displacement)
- k = -305.71 N/m
- x = 0.1 m (displacement at spring-mass rest position)
KE ≈ (3.3 kg * 9.8 m/s^2 * 0.165 m) - (0.5 * -305.71 N/m * (0.1 m)^2)
Calculating the expression gives us the kinetic energy at the spring-mass rest position.
Finally, since kinetic energy is given by:
KE = 0.5 * m * v^2
where v is the speed of the mass, we can rearrange the equation to solve for v:
v = √(2 * KE / m)
Substituting the value of KE we obtained and the given mass value of 3.3 kg into the equation, we can calculate the speed of the mass when it passes through the spring-mass rest position.
Keep in mind to apply the correct units during calculations and round the final answer to an appropriate number of significant figures.