A 2.0 kg ball is placed against a spring of force constant 750N/m, which has been compressed 0.28m. The spring is released and the ball moves along a horizontal frictionless surface and up a ramp. Calculate the...

a) maximum elastic potential energy of the spring (could it be 29.4J, as that's what I've calculated?)
b) maximum velocity of the ball
c) maximum vertical height of the ball on the ramp

Please explain! I want to understand this! Thank you! :)

I will assume that because here is no friction, the ball DOES NOT ROLL.

(1/2) k x^2 = .5 * 750 * .28^2 = 29.4 J check

Ke = (1/2) m v^2 = 29.4
.5 * 2 * v^2 = 29.4
v = 5.42 m/s

m g h = 29.4
2 * 9.81 * h = 29.4
h = 1.5 meters up

Certainly! I'll be happy to explain how to solve these problems.

a) To find the maximum elastic potential energy of the spring, we can use the formula:

Elastic Potential Energy = (1/2) * k * x^2

Where k is the force constant of the spring and x is the compression or extension of the spring. In this case, the compression of the spring is given as 0.28 m, and the force constant is 750 N/m. Plugging these values into the formula:

Elastic Potential Energy = (1/2) * 750 N/m * (0.28 m)^2 = 29.4 J

So yes, your calculation is correct. The maximum elastic potential energy of the spring is indeed 29.4 J.

b) To find the maximum velocity of the ball, we can use the principle of conservation of mechanical energy. As the ball moves up the ramp, the potential energy is converted into kinetic energy. At the maximum height, all the potential energy is converted into kinetic energy.

The initial potential energy of the ball is equal to the elastic potential energy of the spring, which we found to be 29.4 J. At the maximum height, this potential energy is converted into kinetic energy, given by the formula:

Kinetic Energy = (1/2) * m * v^2

where m is the mass of the ball and v is the maximum velocity. Rearranging the formula, we can solve for v:

v = √((2 * Kinetic Energy) / m)

Plugging in the values, we have:

v = √((2 * 29.4 J) / 2 kg) = √(58.8 J / 2 kg) = √29.4 m/s

So the maximum velocity of the ball is approximately 5.42 m/s.

c) To find the maximum vertical height of the ball on the ramp, we can use the principle of conservation of mechanical energy again. At the maximum height, all the potential energy is converted into gravitational potential energy.

Gravitational Potential Energy = m * g * h

where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the maximum height. Rearranging the formula, we can solve for h:

h = Gravitational Potential Energy / (m * g)

Since the gravitational potential energy is equal to the elastic potential energy of the spring, we can use the value we found earlier, which is 29.4 J:

h = 29.4 J / (2 kg * 9.8 m/s^2) = 1.5 m

So the maximum vertical height of the ball on the ramp is 1.5 meters.