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?
a. 1/2 m v^2=1/2 k x^2 where x is the compressed distance, solve for v.
b. Max acceleration? where is max force? Force=kx
c. impluse=changmomentum=mv
To solve this problem, we can use the principles of conservation of energy and momentum.
(a) To find the speed of the box when it passes back through the equilibrium point, we can use the principle of conservation of mechanical energy.
Initially, the box is at rest, so its initial kinetic energy is zero. The spring is compressed by a maximum distance of 38.73 cm, so the potential energy stored in the spring is given by:
Potential energy = (1/2)kx^2
Where k is the spring constant and x is the compression distance.
Potential energy = (1/2) * 4000 N/m * (0.3873 m)^2
≈ 299.4306 Joules
At the equilibrium point, all the potential energy is converted into kinetic energy. Thus, at this point, the kinetic energy is equal to the potential energy:
(1/2)mv^2 = 299.4306 J
Where m is the mass of the box and v is its speed.
Rearranging the equation to solve for v:
v^2 = (2 * 299.4306 J) / 10 kg
v^2 ≈ 59.8861 m^2/s^2
Taking the square root of both sides, we find:
v ≈ √(59.8861) m/s
v ≈ 7.74 m/s
Therefore, the speed of the box when it passes back through the equilibrium point is approximately 7.74 m/s.
(b) To find the maximum magnitude of acceleration of the box, we can use the equation F = ma, where F is the net force acting on the box.
The only force acting on the box is the force exerted by the spring, which is given by Hooke's Law:
F = -kx
Where k is the spring constant and x is the compression distance.
At maximum compression (x = 0.3873 m), the force exerted by the spring is:
F = -4000 N/m * 0.3873 m
F ≈ -1549.2 N
Since mass (m) = 10 kg, we can find the acceleration (a) using F = ma:
-1549.2 N = 10 kg * a
a ≈ -154.92 m/s^2
The negative sign indicates that the acceleration is in the opposite direction to the initial motion of the box.
Therefore, the maximum magnitude of acceleration of the box is approximately 154.92 m/s^2.
(c) To find where the maximum magnitude of acceleration occurs, we need to determine the position where the box reaches its maximum compression.
The maximum compression occurs when the box reverses its direction and starts moving back towards the equilibrium position. At this point, the acceleration is maximum.
Therefore, the maximum magnitude of acceleration occurs at the point of maximum compression, which is 38.73 cm from the equilibrium position.
(d) To find the magnitude of the impulse imparted from the hammer to the box, we can use the principle of conservation of momentum.
Initially, the box is at rest, so its initial momentum is zero. After the impact, the box begins to move with a velocity v.
The change in momentum is given by:
Impulse = Δp = mv
Where m is the mass of the box and v is its velocity.
Since the mass (m) = 10 kg and the velocity (v) = 7.74 m/s (from part a), we can calculate the impulse:
Impulse = 10 kg * 7.74 m/s
Impulse ≈ 77.4 kg m/s
Therefore, the magnitude of the impulse imparted from the hammer to the box is approximately 77.4 kg m/s.
To solve this problem, we need to use the principles of conservation of energy and momentum. Let's break it down step by step:
(a) To find the speed of the box when it passes back through the equilibrium point, we can use the principle of conservation of mechanical energy. At the equilibrium point, all the mechanical energy is in the form of kinetic energy.
1. Determine the potential energy stored in the spring when it is compressed. The potential energy in a spring is given by the formula: PE = (1/2) k x^2, where k is the spring constant and x is the compression or extension of the spring.
In this case, the spring constant k = 4000 N/m and the compression x = 38.73 cm = 0.3873 m.
PE = (1/2) * 4000 N/m * (0.3873 m)^2 = 301.9512 J
2. Since the box is initially at rest, the initial kinetic energy is zero.
3. Using the principle of conservation of mechanical energy: Initial kinetic energy + Initial potential energy = Final kinetic energy
0 + 301.9512 J = (1/2) * mass * final velocity^2
mass = 10 kg (given)
Rearranging the equation, we can solve for the final velocity:
final velocity^2 = (2 * 301.9512 J) / 10 kg
final velocity^2 = 60.39024 m^2/s^2
final velocity = √60.39024 m/s
final velocity = 7.77 m/s (rounded to two decimal places)
Therefore, the speed of the box when it passes back through the equilibrium point is approximately 7.77 m/s.
(b) To find the maximum magnitude of acceleration of the box, we can use Newton's second law, which states that force is equal to mass times acceleration (F = m*a). In this case, the force is provided by the spring.
1. Determine the maximum force exerted by the spring. The maximum force occurs when the spring is fully compressed and is given by Hooke's Law: F = k * x, where k is the spring constant and x is the compression or extension of the spring.
F = 4000 N/m * 0.3873 m = 1549.2 N
2. Use Newton's second law to find the maximum magnitude of acceleration:
F = m * a
a = F / m
a = 1549.2 N / 10 kg
a = 154.92 m/s^2
Therefore, the maximum magnitude of acceleration of the box is 154.92 m/s^2.
(c) To determine where this maximum acceleration occurs, we need to find the position at which the force is maximum. This occurs at the fully compressed state of the spring.
Therefore, the maximum acceleration occurs when the box is fully compressed at the end of the spring.
(d) To find the magnitude of the impulse imparted from the hammer to the box, we can use the principle of conservation of momentum.
1. The impulse, denoted by J, is the change in momentum and is given by J = m * Δv, where m is the mass and Δv is the change in velocity.
2. Since the box was initially at rest, the change in velocity can be calculated as the difference between the final velocity (found in part a) and the initial velocity (which is 0 in this case).
Δv = final velocity - initial velocity
Δv = 7.77 m/s - 0 m/s
Δv = 7.77 m/s
3. Calculate the impulse:
J = m * Δv
J = 10 kg * 7.77 m/s
J = 77.7 N*s
Therefore, the magnitude of the impulse imparted from the hammer to the box is 77.7 N*s.