A 290 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.6 N/cm. The block becomes attached to the spring and compresses the spring 10 cm before momentarily stopping.

(a) While the spring is being compressed, what work is done on the block by the gravitational force on it? J

(b) What work is done on the block by the spring force while the spring is being compressed? J

(c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) m/s

(d) If the speed at impact is doubled, what is the maximum compression of the spring? m

bobpursley, if your not gonna answer the question then don't worry about commenting; this site is not made for you to get on and tell someone its "straightforward"; we come here for help not to be told off. i was hoping to have some help on a similar problem.

thanks

To solve this problem, we will need to use the concepts of work, energy, and Hooke's law. Let's go through each part, step by step:

(a) To find the work done by the gravitational force on the block while the spring is being compressed, we need to determine the change in gravitational potential energy. The formula for gravitational potential energy is given by U = mgh, where U is the gravitational potential energy, m is the mass, g is the acceleration due to gravity, and h is the height.

In this case, the height h is the distance the block falls before it compresses the spring. Since the block initially falls freely, we can use the formula h = 1/2gt^2, where t is the time it takes for the block to reach the spring. We can find t using the equation v = gt, where v is the velocity just before the block reaches the spring. Rearranging the equation, we get t = v/g.

Now, substituting the equation for h and the expression for t into the formula for potential energy, we have U = mg(1/2gt^2). Since we want the work done by the gravitational force, we need to find the change in potential energy, which is U_final - U_initial.

The initial potential energy U_initial = mgh_initial since the block starts at rest, and the final potential energy U_final is zero since the block momentarily stops. Therefore, the work done on the block by the gravitational force is given by:

Work_gravity = U_final - U_initial
= 0 - mgh_initial.

Substituting the given values, Work_gravity = 0 - 290 g (9.8 m/s^2) (h_initial).

(b) The work done on the block by the spring force while it is being compressed can be calculated using the formula for the elastic potential energy of the spring, U_elastic = (1/2) k x^2, where U_elastic is the elastic potential energy, k is the spring constant, and x is the displacement (compression) of the spring.

The work done by the spring force is equal to the negative change in elastic potential energy, as the spring force does negative work when it is being compressed. Therefore, the work done on the block by the spring force is given by:

Work_spring = - (U_elastic_final - U_elastic_initial),
= - [(1/2) k x_final^2 - (1/2) k x_initial^2],
= - (1/2) k (x_final^2 - x_initial^2).

Given that x_initial = 0 cm (since the spring is relaxed initially) and x_final = -10 cm (spring is compressed), we can substitute these values to find the work done by the spring force.

(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 initial mechanical energy (kinetic energy + potential energy) is equal to the final mechanical energy.

Initially, the block has potential energy due to its height, which is given by mgh_initial. When it hits the spring, all this potential energy is converted to kinetic energy, given by (1/2)mv^2, where m is the mass of the block and v is its velocity.

Therefore, we can equate the initial potential energy to the final kinetic energy:

mgh_initial = (1/2)mv^2.

Simplifying and solving for v, we get:

v = sqrt(2gh_initial).

Substituting the given values, we can calculate the speed of the block just before it hits the spring.

(d) If the speed at impact is doubled, we need to find the maximum compression of the spring. To determine this, we'll use the principle of conservation of mechanical energy.

Initially, the block has kinetic energy, given by (1/2)mv^2. When the block hits the spring and comes to a momentary stop, all its kinetic energy is converted to elastic potential energy stored in the compressed spring, given by (1/2)kx^2, where x is the maximum compression.

Therefore, we can equate the initial kinetic energy to the maximum elastic potential energy:

(1/2)mv^2 = (1/2)kx^2.

Simplifying and solving for x, we get:

x = sqrt((mv^2) / k).

Since we're doubling the speed, the new speed is 2v. Substituting this value into the equation, we can find the maximum compression of the spring.

Remember to convert units as needed (e.g., grams to kilograms, cm to meters) to ensure consistent units throughout the calculations.

What is your question here? The problem is rather straightforward.