Kinetic energyAtC=1/2 m v^2
67.67 (your answer)=1/2 * 5 * v^2
calculate velociy v.
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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
67.67 (your answer)=1/2 * 5 * v^2
calculate velociy v.
In this case, the mechanical energy of the block is given by the sum of its kinetic energy (KE) and potential energy (PE) due to the spring and gravity.
At point A, the block has only potential energy due to the spring, which can be calculated as:
PE_A = (1/2) * k * x^2
where k is the spring constant and x is the compression distance of the spring from its equilibrium position, given as 1.5 meters. From the given values, we know that the potential energy at point A is equal to the total work done by the spring:
PE_A = Work_spring = 247.5 J
Substituting the values, we can solve for the spring constant:
(1/2) * k * (1.5^2) = 247.5
k * 2.25 = 495
k = 495 / 2.25
k ≈ 220 N/m
Now, at point B, the block has both kinetic and potential energy. The potential energy due to gravity can be calculated as:
PE_B = m * g * h
where m is the mass of the block, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of point B relative to point A, given as 2.5 meters. From the given values, we know that the kinetic energy at point B is equal to the total work done by gravity:
PE_B = Work_gravity = -122.5 J
Substituting the values, we can solve for the mass of the block:
m * 9.8 * 2.5 = -122.5
m ≈ -122.5 / (9.8 * 2.5)
m ≈ -5 kg
Therefore, the mass of the block is approximately 5 kg.
Now, using the principle of conservation of mechanical energy, we can equate the initial mechanical energy at point A to the final mechanical energy at point C:
KE_A + PE_A = KE_C + PE_C
Since the kinetic energy at point A is given as 247.5 J and the potential energy due to the spring at point A is also 247.5 J, we can rewrite the equation as:
247.5 J + 247.5 J = KE_C + PE_C
495 J = KE_C + PE_C
To determine the potential energy at point C, we need to consider that the block has reached a height of 2.5 meters above point A, and the potential energy due to gravity is given by:
PE_C = m * g * h
Since the block moves beyond point B by a distance of 2.6 meters, we can calculate the velocity at point B using the work done by the horizontal surfaces:
Work_horizontal = -57.33 J
The work done by friction is equal to the change in kinetic energy:
Work_horizontal = KE_B - KE_A
-57.33 J = KE_B - 247.5 J
KE_B = -57.33 J + 247.5 J
KE_B ≈ 190.17 J
Now, using the given kinetic energy at point B, we can find the velocity at point B:
KE_B = (1/2) * m * v_B^2
190.17 J = (1/2) * 5 kg * v_B^2
v_B ≈ sqrt(190.17 J / (2 * 5 kg))
v_B ≈ sqrt(19.017 m^2/s^2)
v_B ≈ 4.361 m/s
Now that we have the velocity at point B, we can calculate the velocity at point C using the conservation of mechanical energy equation:
495 J = KE_C + m * g * h
KE_C = 495 J - m * g * h
KE_C = 495 J - 5 kg * 9.8 m/s^2 * 2.5 m
KE_C ≈ 495 J - 122.5 J
KE_C ≈ 372.5 J
Finally, we can find the velocity at point C:
KE_C = (1/2) * m * v_C^2
372.5 J = (1/2) * 5 kg * v_C^2
v_C ≈ sqrt(372.5 J / (2 * 5 kg))
v_C ≈ sqrt(37.25 m^2/s^2)
v_C ≈ 6.103 m/s
Therefore, the speed of the block when it reaches point C is approximately 6.103 m/s.