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 or 5.2 that is not the answer

User Icon for Anonymous Anonymous answered
3 years ago

67.67 = (1/2) m v^2 = (1/2)(5) v^2

well, I get 5.2 all right. I think there is probably a typo somewhere.

User Icon for Explain Bot Explain Bot answered
11 months ago

To calculate the speed of the block when it reaches point C, we will use the principle of conservation of mechanical energy.

The mechanical energy of the block is the sum of its kinetic energy (KE) and potential energy (PE). In this case, the potential energy is the gravitational potential energy given by PE = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height above the reference point.

Let's break down the problem into different stages:

Stage 1: Block compressed against the spring (before moving)
- The block is not moving initially, so its kinetic energy at this stage is 0 joules.
- The potential energy due to compression of the spring is given by PE = (1/2)kx^2, where k is the spring constant and x is the compression distance.
- Given that the spring constant k is 220 N/m and the compression distance x is 1.5 m, we can calculate the potential energy by substituting these values into the equation.

Stage 2: Block moves from the initial position to point A (along a horizontal surface)
- The total work done by the spring on the block is equal to the change in mechanical energy, which is the difference between its potential energy at stage 1 and the kinetic energy at stage 2.
- The work done by the spring is equal to the negative of the change in potential energy, as the spring force opposes the motion of the block.
- Given that the total work done by the spring is 247.5 joules and the potential energy at stage 1 is already calculated, we can find the kinetic energy at stage 2.

Stage 3: Block moves from point A to point B (along an incline)
- The height difference between point A and point B is given as 2.5 meters.
- The total work done by the Earth's gravity on the block is equal to the change in mechanical energy, which is the difference between its potential energy at stage 2 and the kinetic energy at stage 3.
- The work done by gravity is equal to the negative of the change in potential energy, as gravity does negative work when the block moves up against it.
- Given that the total work done by gravity is -122.5 joules and the kinetic energy at stage 2 is already calculated, we can find the kinetic energy at stage 3.

Stage 4: Block moves from point B to point C (along a horizontal surface with significant friction)
- The total work done by the horizontal surfaces due to friction on the block is equal to the change in mechanical energy, which is the difference between its kinetic energy at stage 3 and the kinetic energy at stage 4.
- The work done by friction is equal to the negative of the change in kinetic energy, as friction opposes the motion of the block.
- Given that the total work done by friction is -57.33 joules and the kinetic energy at stage 3 is already calculated, we can find the kinetic energy at stage 4.

Stage 5: Block moves past point C and continues with negligible friction
- Since the region past point C has negligible friction and no external work is done on the block, the mechanical energy remains constant.
- The kinetic energy at stage 4 is equal to the kinetic energy at stage 5.

Stage 6: Block reaches point C
- The kinetic energy at stage 5 is the maximum potential energy at point C.
- The block is moving at constant speed at this point, so its kinetic energy is equal to the maximum potential energy.
- Given that the kinetic energy at stage 5 is 67.67 joules, we can calculate the speed of the block at point C.

By using these calculations and the intermediate values provided in the problem, you can find the speed of the block when it reaches point C.