A 1.0 kg mass that can move along the x-axis experiences the potential energy U = (x²−x) J, where x is in m. The mass has velocity v_x= 3.0 m/s at position x = 1.0 m. At what position has it slowed to 1.0 m/s?

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Hi, everyone. In this practice problem, we are being asked to determine the object's velocity. When it reaches the position of S equals to eight m, we were given a 2.5 kg object which is going to be initially at rest at position of S equals to five m. On a potential energy diagram given by the equation of us equals to negative to S cube plus four S squared J. We were being asked to determine the object's velocity when it reaches position S E close to eight m. And the options given are a 22.2 m per second. B 49.4 m per second, C 21.2 m per second and D 29.4 m per second. So in order for us to actually do this problem, we want to consider the object as a particle, we want to then apply the principle of conservation of energy, which by the principle of conservation of energy, we will have the summation of the initial potential and kinetic energy to be equals to the submission of the final potential and kinetic energy. So I'm gonna represent potential energy as U and kinetic energy as K and the subscript of I as the initial and F as the final. So the initial potential energy plus the initial kinetic energy will be equals to the final potential energy with the subscript of F plus the final kinetic energy, which is K F. Now, we can actually substitute everything in our, substitute our equations in. And the, the equation for the potential energy is given by the problem statement which is us equals to negative two S cube plus four S squared. So for the initial potential energy U I then that will be open parenthesis, negative two si cube plus four si squared. And that will be the initial potential energy we wanna plus that or add that with the kinetic initial kinetic energy which is going to be half M V I squared and V I is going to be the initial velocity. The, the final potential energy will follow the same formula or equation which is negative two S F cube plus four S F squared I wanna plus that with the kinetic energy which is half M V F squared for the final velocity. And what's being asked for the problem state from the problem statement is the object's final velocity when it reaches the position of S equals eight m, which is going to be our V F here. So in this case, our si is going to be five m which is given in the problem statement, our S F is going to be eight m and our V I is going to be zero m per second because our object initially starts at rest. So we want to substitute dose into our uh formula that we have here or our expression. So this will then be negative two si Q will be five m cube plus four si squared, which is four multiplied by a five m squared plus half M V I squared, which is half the mass is given to be 2.5 kg. Uh V I squared is going to be 00 m per second squared, which essentially this whole term for the initial kinetic energy will equals to zero. And all of that will then be equal to the final condition or final energy which is negative to S F cube, which is going to be eight m cube plus four S F squared, which is going to be eight m squared. We wanna add that with half of the mass which is 2.5 kg and the final velocity which is being asked from there doing our calculation, we will then get 618 to be equals to half of 2.5 kg multiplied by V F squared just like. So rearranging this and then recalculating V F squared will then equals to B 494.4 m squared per second squared. Taking the square root of that. Then R V F will then equals to the square root of 494.4 m squared divided by S squared. And that will give us a vinyl V F value of 22.2 m per second, which will be the object's velocity when it reaches the position of S equals to eight m, which is going to correspond to option a in our answer choices. So answer a with the objects velocity of 22.2 m per second at position S equals eight m is going to be the answer to this particular practice problem. So if you guys still have any sort of confusion on this particular problem, please make sure to check out our other lesson videos on similar topic and that will be all for this one. Thank you.

A 1.0 kg mass that can move along the x-axis experiences the potential energy U = (x²−x) J, where x is in m. The mass has velocity v_x= 3.0 m/s at position x = 1.0 m. At what position has it slowed to 1.0 m/s?

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A 1.0 kg mass that can move along the x-axis experiences the potential energy U = (x²−x) J, where x is in m. The mass has velocity vx = 3.0 m/s at position x = 1.0 m. At what position has it slowed to 1.0 m/s?

Verified video answer for a similar problem:

Key Concepts

Potential Energy

Kinetic Energy

Conservation of Mechanical Energy

A 1.0 kg mass that can move along the x-axis experiences the potential energy U = (x²−x) J, where x is in m. The mass has velocity v_x= 3.0 m/s at position x = 1.0 m. At what position has it slowed to 1.0 m/s?