Final answer:
The speed of the block just before it hits the spring is 10 m/s. The spring has been compressed 1.82574 m when the block comes to rest. The block reaches its maximum speed after the collision with the spring at the location where the block first comes to rest, and the maximum speed is 10 m/s.
Explanation:
This question is related to physical properties such as mass, acceleration due to gravity, angles, and kinetic energy involving a block sliding down a frictionless plane, a spring, and conservation of mechanical energy.
a. Speed of the block before hitting the spring: It can be established using the principle of conservation of energy. Here, potential energy is converted into kinetic energy. Potential energy is m*g*h and as h of the block is d*sin(θ)=5*0.5=2.5m, potential energy = 10*10*2.5 = 250 J. If all potential energy becomes kinetic energy, then 1/2*m*v^2 = 250, solving for v, we get v = sqrt(500/10)=10 m/s.
b. Distance the spring compresses: Again, the law of conservation of energy applies. Here, kinetic energy is converted into spring potential energy (1/2*K*x^2). From the earlier calculation, the kinetic energy upon impact is 250 J. So, 250 = 1/2 * 150 * x^2. Solving for x gives x = 1.82574 m.
c. Location of maximum block speed post-collision: This would be at the point where the block first comes to rest after striking the spring since it goes from moving to stopped.
d. Maximum speed of the block after the collision: Since the mechanical energy is conserved and no other forces are acting on the block, the velocity before and after interaction with the spring should be the same, i.e., 10 m/s.
