A box with a mass of m = 7.50 kg is set on top of a spring with a spring constant

of k = 500 N/m. The box is pushed down 30.0 cm and released.
(a) What is the maximum height the box reaches after being launched by the
spring?
(b) What is the instantaneous power delivered to the box by the spring when it
is only compressed by 10.0 cm?

To solve this problem, we will use the principles of potential energy and work done by the spring.

(a) To find the maximum height reached by the box, we can use the conservation of mechanical energy. The initial energy stored in the spring is equal to the maximum potential energy reached by the box.

The potential energy stored in the spring is given by the equation:
Potential energy (PE) = (1/2) * k * x^2

Where k is the spring constant and x is the displacement from the equilibrium position.

The initial potential energy stored in the spring is equal to the gravitational potential energy at the maximum height:
PE_initial = m * g * h_max

Where m is the mass of the box, g is the acceleration due to gravity, and h_max is the maximum height reached by the box.

Setting the initial potential energy equal to the final potential energy, we have:
(1/2) * k * x^2 = m * g * h_max

Substituting the given values, we have:
(1/2) * 500 N/m * (0.30 m)^2 = 7.50 kg * 9.8 m/s^2 * h_max

Simplifying the equation, we can solve for h_max:
h_max = ((1/2) * 500 N/m * (0.30 m)^2) / (7.50 kg * 9.8 m/s^2)

Calculating the value, h_max ≈ 0.915 m

Therefore, the maximum height the box reaches after being launched by the spring is approximately 0.915 meters.

(b) To find the instantaneous power delivered to the box by the spring when it is compressed by 10.0 cm, we need to calculate the force exerted by the spring at that point.

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.

The instantaneous power delivered by the spring is given by the equation:
Power = F * v

Where Power is the instantaneous power, F is the force exerted by the spring, and v is the velocity of the box.

At the maximum compression of 10.0 cm (0.10 m), the force exerted by the spring can be calculated as:
F = -k * x = -500 N/m * 0.10 m

The velocity of the box at that point can be determined using energy conservation:
Initial potential energy = Final kinetic energy
(1/2) * k * x^2 = (1/2) * m * v^2

Substituting the given values, we have:
(1/2) * 500 N/m * (0.10 m)^2 = (1/2) * 7.50 kg * v^2

Solving for velocity, we have:
v = sqrt((500 N/m * (0.10 m)^2) / (7.50 kg))

Calculating the value, v ≈ 0.79 m/s

Now, we can calculate the instantaneous power as:
Power = F * v = (-500 N/m * 0.10 m) * 0.79 m/s

Calculating the value, Power ≈ -39.5 W

Note that the negative sign indicates that the power is being delivered to the box by the spring.

Therefore, the instantaneous power delivered to the box by the spring when it is compressed by 10.0 cm is approximately -39.5 Watts.