A vertical spring (ignore its mass), whose spring constant is 980 N/m, is attached to a table and is compressed 0.160 m.
(a) What speed can it give to a 0.400 kg ball when released?
(b) How high above its original position (spring compressed) will the ball fly?
V=12
To answer these questions, we can use the concepts of the conservation of mechanical energy and Hooke's Law.
(a) The first step is to find the potential energy stored in the compressed spring. The formula for the potential energy stored in a spring is given by:
Potential Energy (PE) = (1/2) * k * x^2
Where k is the spring constant and x is the displacement from the equilibrium position.
In this case, the spring constant (k) is given as 980 N/m and the displacement (x) is given as 0.160 m. Plugging these values into the formula, we get:
PE = (1/2) * 980 N/m * (0.160 m)^2 = 12.544 J (joules)
Since energy is conserved, the potential energy stored in the spring will be converted entirely into kinetic energy when the ball is released. Therefore, the kinetic energy of the ball can be calculated as:
Kinetic Energy (KE) = Potential Energy (PE)
KE = 12.544 J
The kinetic energy of an object is given by the formula:
KE = (1/2) * m * v^2
Where m is the mass of the object and v is its velocity.
Rearranging the formula, we can solve for the velocity (v):
v^2 = (2 * KE) / m
v = √[(2 * KE) / m]
Plugging in the values, we get:
v = √[(2 * 12.544 J) / 0.400 kg] ≈ 5.60 m/s
Therefore, the speed the ball can give when released is approximately 5.60 m/s.
(b) To find how high the ball will fly above its original position, we can use the conservation of mechanical energy. When the ball is at its highest point, all of its initial kinetic energy will be converted into potential energy due to gravity.
The potential energy of an object at height h can be calculated as:
Potential Energy (PE) = m * g * h
Where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
Since the kinetic energy is converted into potential energy at the highest point, we can equate the kinetic energy at the release point to the potential energy at the highest point:
KE = PE
(1/2) * m * v^2 = m * g * h
Simplifying the equation and solving for h, we get:
h = (v^2) / (2*g)
Plugging in the values, we get:
h = (5.60 m/s)^2 / (2 * 9.8 m/s^2) ≈ 1.79 m
Therefore, the ball will fly approximately 1.79 meters above its original position.
do
72=9D
u should get
8=D