A 10-kg box is at rest on a horizontal frictionless surface, at the end of an unstretched horizontal spring with constant k = 4000 N/m. The mass is then quickly struck with a hammer that imparts an impulse along the horizontal direction, compressing the spring a maximum distance of 38.73 cm. a) What is the speed of the box when it passes back through the equilibrium point? (b) What is the maximum magnitude of acceleration of the box and (c) where does it occur? (d) What is the magnitude of the impulse that was imparted from the hammer to the box?

To find the answers to these questions, we can use the concept of conservation of mechanical energy and the equations of motion. Let's break down the problem step by step.

(a) What is the speed of the box when it passes back through the equilibrium point?

To find the speed of the box, we need to first find the spring potential energy at the maximum compression point. The formula for spring potential energy is given by:

PE_spring = (1/2) * k * x^2

where k is the spring constant and x is the displacement of the spring from equilibrium. In this case, x = 0.3873 m (since it is given in centimeters).

Plugging the values into the formula:

PE_spring = (1/2) * 4000 N/m * (0.3873 m)^2 = 299.9248 J

Since the box is at rest initially, the initial kinetic energy is zero. Therefore, the total mechanical energy at the maximum compression point should be equal to the spring potential energy:

Total mechanical energy = PE_spring = 299.9248 J

At the equilibrium point, all of the potential energy is converted to kinetic energy:

Total mechanical energy = Kinetic energy

So, we can write:

(1/2) * m * v_eq^2 = 299.9248 J

where m is the mass of the box and v_eq is the speed of the box at the equilibrium point that we need to find.

Plugging in the values:

(1/2) * 10 kg * v_eq^2 = 299.9248 J

Simplifying:

v_eq^2 = 599.8496 J / kg
v_eq = √(599.8496 J / kg)
v_eq = 7.745 m/s

Therefore, the speed of the box when it passes back through the equilibrium point is approximately 7.745 m/s.

(b) What is the maximum magnitude of acceleration of the box?

To find the maximum magnitude of acceleration, we use the equation:

a_max = k * x_max / m

where k is the spring constant, x_max is the maximum compression distance, and m is the mass of the box.

Plugging in the values:

a_max = (4000 N/m) * (0.3873 m) / 10 kg
a_max ≈ 15.49 m/s^2

Therefore, the maximum magnitude of acceleration of the box is approximately 15.49 m/s^2.

(c) Where does it occur?

The maximum magnitude of acceleration occurs at the maximum compression point of the spring.

(d) What is the magnitude of the impulse that was imparted from the hammer to the box?

Impulse is defined as the change in momentum of an object. In this case, the impulse imparted from the hammer to the box can be calculated by finding the change in velocity before and after the hammer strike and multiplying it by the mass of the box.

Since the box is initially at rest, the change in velocity is equal to the final velocity:

Impulse = m * Δv = m * (v_eq - 0)

Plugging in the values:

Impulse = 10 kg * 7.745 m/s
Impulse = 77.45 kg⋅m/s

Therefore, the magnitude of the impulse imparted from the hammer to the box is 77.45 kg⋅m/s.