Problem 4:
A 250 g block is dropped onto a relaxed vertical spring that has a spring constant of k= 2.5 N/cm. The block becomes attached to the spring and compresses the spring 12 cm before momentarily stopping. While the spring is being compressed,
(a) what work is done on the block by the gravitational force on it?
(b) what work is done on the block by the spring force?
(c) what is the speed of the block just before it hits the spring? (neglect friction).
(a) The work done on the block by the gravitational force is equal to the change in the gravitational potential energy of the block. The gravitational potential energy of the block is equal to mgh, where m is the mass of the block (250 g), g is the acceleration due to gravity (9.8 m/s2), and h is the height from which the block is dropped. Since the block is dropped from rest, the change in gravitational potential energy is equal to mgh. Therefore, the work done on the block by the gravitational force is equal to 250 g x 9.8 m/s2 x 12 cm = 294 J.
(b) The work done on the block by the spring force is equal to the change in the elastic potential energy of the spring. The elastic potential energy of the spring is equal to 1/2 kx2, where k is the spring constant (2.5 N/cm) and x is the compression of the spring (12 cm). Therefore, the work done on the block by the spring force is equal to 1/2 x 2.5 N/cm x (12 cm)2 = 180 J.
(c) The speed of the block just before it hits the spring can be calculated using the equation v2 = 2gh, where v is the speed of the block, g is the acceleration due to gravity (9.8 m/s2), and h is the height from which the block is dropped (12 cm). Therefore, the speed of the block just before it hits the spring is equal to √2 x 9.8 m/s2 x 12 cm = 17.3 m/s.
To solve this problem, we need to consider the work done by the gravitational force and the work done by the spring force. We also need to determine the speed of the block just before it hits the spring.
(a) To find the work done by the gravitational force, we can use the formula:
Work = Force * distance * cos(theta)
In this case, the force due to gravity is the weight of the block, which can be calculated using the formula:
Weight = mass * acceleration due to gravity
The acceleration due to gravity is approximately 9.8 m/s^2. So, the weight of the block is:
Weight = 0.25 kg * 9.8 m/s^2 = 2.45 N
The distance over which the gravitational force acts is the height from which the block is dropped. This information is not given in the problem, so we will assume it to be given in the future.
(b) The work done by the spring force can be calculated using the formula:
Work = (1/2) * k * x^2
Where k is the spring constant and x is the displacement of the spring from its equilibrium position. In this case, the spring is compressed by 12 cm, which is equivalent to 0.12 meters. Therefore, the work done by the spring force is:
Work = (1/2) * 2.5 N/cm * (0.12 m)^2
(c) To find the speed of the block just before it hits the spring, we can use the principle of conservation of mechanical energy. The potential energy when the block is at rest on the spring is converted into kinetic energy just before it hits the spring.
The potential energy of the spring can be calculated using the equation:
Potential Energy = (1/2) * k * x^2
Where k is the spring constant and x is the compression of the spring. In this case, the potential energy of the spring is:
Potential Energy = (1/2) * 2.5 N/cm * (0.12 m)^2
The potential energy is converted into kinetic energy, given by the equation:
Kinetic Energy = (1/2) * mass * velocity^2
Where mass is the mass of the block and velocity is the speed of the block just before it hits the spring.
Equating the potential energy to the kinetic energy, we have:
(1/2) * 2.5 N/cm * (0.12 m)^2 = (1/2) * 0.25 kg * velocity^2
Simplifying the equation, we can solve for the velocity:
velocity^2 = (2.5 N/cm * (0.12 m)^2) / (0.25 kg)
Once we find the value of velocity, we can take the square root to get the speed of the block just before it hits the spring.
Please provide the height from which the block is dropped to proceed with the calculations.
To solve this problem, we need to apply the principles of work and energy conservation. Let's break down each part of the question:
(a) To determine the work done on the block by the gravitational force, we need to find the change in gravitational potential energy. The formula for gravitational potential energy is given by:
Potential energy (PE) = mass (m) * gravity (g) * height (h)
In this case, the mass of the block (m) is 250 g, which is equivalent to 0.250 kg. The height (h) is the distance the block falls before hitting the spring, which we'll calculate later.
(b) The work done on the block by the spring force can be found using Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed. The formula is:
Force (F) = spring constant (k) * displacement (x)
where the displacement (x) is the distance the spring is compressed, which in this case is given as 12 cm.
(c) To determine the speed of the block just before it hits the spring, we need to use the principle of conservation of mechanical energy. This principle states that the total mechanical energy of an object is conserved in the absence of external forces (like friction).
The total mechanical energy is the sum of potential energy (due to gravity) and kinetic energy. Since the block is initially at rest, the potential energy is zero. Therefore, the total mechanical energy is equal to the kinetic energy just before hitting the spring.
The formula for kinetic energy is:
Kinetic energy (KE) = 0.5 * mass (m) * velocity^2
where the mass (m) is 0.250 kg.
Now that we have outlined the steps to solve each part of the problem, let's calculate the answers:
(a) Work done by the gravitational force:
PE = m * g * h
Since h is the distance the block falls before hitting the spring, we need to calculate it. The block is initially at rest, so we can use the equation for distance (d) traveled when an object is accelerated:
d = 0.5 * g * t^2
In this case, the initial velocity (vo) is zero. Solving for time (t) gives us:
t = sqrt((2 * d) / g)
where d is the distance the block falls. Plugging in the values:
t = sqrt((2 * h) / g)
Substituting this value of t into the equation for height (h):
h = 0.5 * g * t^2
Now we can calculate the height:
h = 0.5 * 9.8 * (sqrt((2 * 12) / 9.8))^2
(b) Work done by the spring force:
Work (W) = F * x
Substituting the given values:
W = k * x
W = 2.5 N/cm * 12 cm
(c) To find the speed just before the block hits the spring, we need to use the conservation of mechanical energy:
Total Mechanical Energy (TME) = Potential Energy (PE) + Kinetic Energy (KE)
PE = 0
KE = 0.5 * m * v^2
Since TME is conserved:
TME = KE = 0.5 * m * v^2
Substituting the values:
0.5 * m * v^2 = 0.5 * m * (0 m/s)^2