A block of mass m is dropped onto a relaxed spring constant k. The block becomes attached to the spring and compresses the spring a distance d before momentarily stopping. (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)

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

(b) While the spring is being compressed, what work is done on the block by the spring force?
W =

(c) What is the speed of the block just before it hits the spring?
v =

To answer these questions, we need to apply the concepts of work and energy conservation.

(a) To determine the work done on the block by the gravitational force, we can use the equation for work: W = F*d*cosθ, where F is the force applied, d is the displacement, and θ is the angle between the force and displacement vectors.

In this case, the gravitational force acting on the block is mg, and the displacement is the distance the block falls, which is determined by the height it was dropped from. The angle between the force and displacement vectors is 0 degrees since they are in the same direction.

Therefore, the work done by the gravitational force is:

W = mg*d*cos0
W = mgd

(b) The work done on the block by the spring force can be calculated using the formula W = (1/2)k*x^2, where k is the spring constant and x is the distance the spring is compressed.

In this case, the work done by the spring force is given by:

W = (1/2)k*d^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 total mechanical energy of the system (block + spring) is conserved, which means the initial potential energy of the block when it was dropped is converted into kinetic energy just before it hits the spring.

The potential energy of the block when it was dropped is mgh, where h is the height it was dropped from.

The kinetic energy just before it hits the spring is (1/2)mv^2.

By conserving energy, we have:

mgh = (1/2)mv^2

Simplifying the equation, we find:

gh = (1/2)v^2

Solving for v:

v = sqrt(2gh)

Therefore, the speed of the block just before it hits the spring is given by the square root of 2 times the product of g and h.