Hung vertically, a massless spring extends by 3.00 cm when a mass of 992.0 g is attached to its lower end. The same mass and spring are then placed apart on a table. The spring is fixed in place and then the mass is given a velocity of 0.800 m/s towards the spring.
a) What is the maximum compression of the spring when the mass runs into it?
b) After compressing the spring the mass will rebound. What is its velocity just as it leaves contact with the spring?
a, find k. k= mg/x (in change grams to kg, cm to m).
then, 1/2 mv^2= 1/2 k x^2
solve for x, the max compression.
b. it will rebound with a velocity of .800m/s, of course...if the spring is massless..
thanks!
To solve this problem, we can apply the principle of conservation of mechanical energy. The total mechanical energy of the system (spring + mass) remains constant throughout the motion.
a) To find the maximum compression of the spring when the mass runs into it, we need to determine the potential energy stored in the spring when it is fully compressed.
1. Calculate the spring constant (k):
We know that the massless spring extends by 3.00 cm when a mass of 992.0 g (0.992 kg) is attached to it. The force exerted by the spring is given by Hooke's Law, 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.
Since the spring is in equilibrium when there is no mass attached to it, the force exerted by the spring is equal to the weight of the mass attached to it. So we have:
k * (0.03 m) = m * g
k * (0.03 m) = (0.992 kg) * (9.8 m/s^2)
k = (0.992 kg * 9.8 m/s^2) / 0.03 m
k = 323.7333 N/m (approximately)
2. Determine the potential energy stored in the spring at maximum compression:
The potential energy stored in a spring is given by the equation U = (1/2) * k * x^2, where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
When the spring is maximally compressed, x will be equal to the maximum compression of the spring. So we have:
U = (1/2) * (323.7333 N/m) * (0.03 m)^2
U = 0.145680 Nm (approximately)
b) To find the velocity of the mass just as it leaves contact with the spring, we can again use the principle of conservation of mechanical energy.
At the maximum compression of the spring, all the potential energy stored in the spring is converted into kinetic energy of the mass. Therefore, the kinetic energy of the mass just as it leaves contact with the spring is equal to the potential energy stored in the spring.
1. Calculate the kinetic energy of the mass:
The kinetic energy of an object is given by the equation K = (1/2) * m * v^2, where K is the kinetic energy, m is the mass of the object, and v is the velocity of the object.
So we have:
0.145680 Nm = (1/2) * (0.992 kg) * v^2
v^2 = (2 * 0.145680 Nm) / 0.992 kg
v^2 = 0.29435 m^2/s^2
Taking the square root of both sides, we find:
v ≈ 0.542 m/s (approximately)
Therefore, the velocity of the mass just as it leaves contact with the spring is approximately 0.542 m/s.