Need help with a practice problem. Image included ### Physics Problem: Block on a Frictionless Ramp#### Problem Statement:A block with mass 3.5 kg is on a frictionless ramp with a spring at the bottom that has a spring constant of 460 N/m. The angle of the ramp is 39°.#### Diagram Explanation:The diagram shows a ramp inclined at an angle θ (39°) from the horizontal. A block of mass "m" is initially positioned at a distance "d" from the spring at the bottom of the ramp.#### Questions:1. **Compression of the Spring:** - The block slides down the ramp a distance of 67 cm before hitting the spring. Calculate how far in centimeters the spring is compressed when the block comes to a momentary rest.2. **Speed of the Block:** - After the block comes to rest, the spring pushes the block back up the ramp. Determine the speed of the block in meters per second as it moves up the ramp after it comes off the spring.3. **Change in Gravitational Potential Energy:** - Calculate the change in gravitational potential energy (in joules) between the original position of the block at the top of the ramp and its position when the spring is fully compressed.#### Key Concepts:- **Spring Compression:** Use the principle of conservation of energy to find the compression distance.- **Kinematics and Energy:** Consider gravitational potential energy and spring potential energy.- **Potential Energy Change:** Calculate using the formula \( \Delta U = mgh \).#### Relevant Equations:- Hooke’s Law for springs: \( F = -kx \)- Potential energy of spring: \( U = \frac{1}{2}kx^2 \)- Gravitational potential energy: \( U = mgh \)- Energy conservation: \( \text{Initial Potential Energy} = \text{Spring Potential Energy} \) + \text{Kinetic Energy (if any)} These concepts will help analyze the motion and forces involved in the problem.

Given,mass of the block, spring constant, the angle of the ramp, distance covered by block while sliding down the rampThe force acting on the block along the inclined plane,Now from the free-body diagram of the block,The velocity of the block when it slides 67 cm down the ramp,from the third equation of motion,from the conservation of energy the kinetic energy of the block will be converted into the potential energy of the spring,Hence the spring will get compressed up to 25 cm only.Part B The ramp is frictionless so the block will go back at the same speed at which it collides with the spring, with the velocity ofPart CIn this case, the change in the gravitational potential energy will be the negative value of kinetic energy,Hence the negative value of kinetic energy will be -14.4 J.a)Hence the spring will get compressed up to 25 cm only.b)Velocity , v= 2.87 m/sc)change in the gravitational potential energy will be 14.4 J