How Do You Calculate the Velocity of a 3.00kg Block at x=2.00m?

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To calculate the velocity of a 3.00 kg block at x=2.00 m, the conservation of mechanical energy principle is applied, considering both kinetic and potential energy. The potential energy function is given as U=(-x^2)-4x+5, which results in a potential energy of 7 J at x=0. The block's initial kinetic energy at x=0 is based on its speed of 5.00 m/s, leading to a total energy that must be conserved. The discrepancy in calculated velocities stems from different interpretations of the force's direction and the change in potential energy. Ultimately, the correct answer is confirmed to be 5.74 m/s, highlighting the importance of accurately applying energy conservation principles.
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Find velocity...

A 3.00Kg block at x=0 is moving to the right at 5.00m/s.
A horizontal force acts on the block. Its potential energy funtion is U=(-x^2)-4x+5
find the speed of the block at x=2.00m (surface is smooth)

ok the answer is already given by prof. answer is 5.74m/s but I want to know how to solve this. I used kinetic and potential energy equation for this problem but I got 4.52m/s also I used changing kinetic and potential energy but I got 8.50m/s...I tried other things too but I got wrong... I'm stuck with this...
 
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The answer is either 4.51 m/s or 5.45 m/s, depending on whether the force is opposite or with the direction of motion. If the answer is actually 5.74 m/s, then the question is wrong.

Clearly the function gives a potential energy of 7 J, so the block either gains or loses 7 J of KE (depending on the intended meaning of the negative quantity).

Edit: I noted rock.freak's answer, below, and then slapped my head. Only the change in PE of the system matters. Since the +5 exists at zero, and at x=2, then then change in KE is 12 J.

Awkwardly worded question.
 
Last edited:
U=-x^2-4x+5

at x=0..what is U? Then use the law of conservation of mechanical energy.
 

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