A yellow 5.0 kg block slides over a surface with negligible friction (except in a higher-friction region indicated by a red and black patch between points B and C in the diagrams above).The upper diagram shows the situation before the block has started moving. The block is touching an ideal spring with a spring constant k of 220 N/m. The spring has been compressed a distance of 1.5 meters from its equilibrium position.The lower diagram shows the block shortly after it has been released and it has moved to location A. Shortly after this it will slide up the incline (also with negligible friction) and reach point B at a height of 2.5 meters higher than point A. It will then slide to the right, beyond point B, over the only region with significant friction (red and black in the diagram) before reaching point C. The block continues to slide past point C with negligible friction. The region with significant friction is 2.6 meters long, and the coefficient of friction between the surfaces is 0.45.

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What you also know:
- The total work that the spring does on the block between the starting position and the time the block has reached the far end at point C is (247.5 joules)
- The kinetic energy of the block when it is at point A is (247.5 joules)
- The total work that the Earth does on the block (via gravity) between the starting position and the time the block has reached the far end at point C is (-122.5 joules)
- The kinetic energy of the block when it is at point B is (125 joules)
- The total work that horizontal surfaces do on the block (via friction) between the starting position and the time the block has reached the far end at point C is (-57.33 joules)
- The kinetic energy of the block when it reaches point C is (67.67 joules)
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THE QUESTION TO ANSWER (USE INFO ABOVE TO ANSWER)
1) Calculate the speed of the block when it reaches point C.

NOTE: A lot is given to you I just need help on one question which is finding the speed so please solve because I'm stuck figuring it out...much appreciated also if you get 5.20269 that is not the answer for this question

3 answers

Kinetic energyAtC=1/2 m v^2
67.67 (your answer)=1/2 * 5 * v^2
calculate velociy v.
so the problem is your KE at point C if 5.2m/s is incorrect.
so what would be the answer then? I'm confuse on what your trying to say