In question 7, you will need to upload your scratch paper for this problem. U (x) x = 0 x = x0 Q Potential energy U(x) The graph above shows the potential energy of an object as a function of position, x. The object begins at the point P, a distance of xo = 0.79 meters from the origin. The object needs to get to the point Q, as shown in the diagram. You do not need to know the exact location of point Q to solve this problem, but the x-coordinate of point Q is greater than 5 xo. Between the points P and Q, the potential energy of the object is given by: U(x) = -a x² + bx – c [That says U(x) = - a x^2 + b x - c ] Where a = 0.43 kg/s? , b = 2.3 kg m/s², and c=4.2 joules .
In question 7, you will need to upload your scratch paper for this problem. U (x) x = 0 x = x0 Q Potential energy U(x) The graph above shows the potential energy of an object as a function of position, x. The object begins at the point P, a distance of xo = 0.79 meters from the origin. The object needs to get to the point Q, as shown in the diagram. You do not need to know the exact location of point Q to solve this problem, but the x-coordinate of point Q is greater than 5 xo. Between the points P and Q, the potential energy of the object is given by: U(x) = -a x² + bx – c [That says U(x) = - a x^2 + b x - c ] Where a = 0.43 kg/s? , b = 2.3 kg m/s², and c=4.2 joules .
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Can you please answer this
![In question 7, you will need to upload your scratch paper for this problem.
U(x)
x = 0
x = x0
Q
Potential energy U(x)
The graph above shows the potential energy of an object as a function of position, x. The object
begins at the point P, a distance of xo = 0.79 meters from the origin. The object needs to get to the
point Q, as shown in the diagram. You do not need to know the exact location of point Q to solve
this problem, but the x-coordinate of point Q is greater than 5 xo-
Between the points P and Q, the potential energy of the object is given by:
U(2) =
= -a x + b x
C
[That says U(x) = - a x^2 + b x - c]
Where a
= 0.43 kg/s² , b = 2.3 kg m/s² , and c=4.2 joules .](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8cd02e14-2925-4f0b-b5d5-f979054af505%2F60b5057b-d2c5-414f-8ba1-f15ede1e8b4e%2Fbrq2ydq_processed.png&w=3840&q=75)
Transcribed Image Text:In question 7, you will need to upload your scratch paper for this problem.
U(x)
x = 0
x = x0
Q
Potential energy U(x)
The graph above shows the potential energy of an object as a function of position, x. The object
begins at the point P, a distance of xo = 0.79 meters from the origin. The object needs to get to the
point Q, as shown in the diagram. You do not need to know the exact location of point Q to solve
this problem, but the x-coordinate of point Q is greater than 5 xo-
Between the points P and Q, the potential energy of the object is given by:
U(2) =
= -a x + b x
C
[That says U(x) = - a x^2 + b x - c]
Where a
= 0.43 kg/s² , b = 2.3 kg m/s² , and c=4.2 joules .
Transcribed Image Text:The object must get from point P to point Q using only the energy it has when it starts at point P,
without a motor or external forces. The is no friction and there are no forces aside from the
conservative force associated with this potential energy.
Calculate the minimum kinetic energy (in joules) the object must have when it is at point P in order to
reach point Q.
Three notes about this problem:
1) This problem is similar to one that you have done involving two planets and escape
velocity, but the potential energy function is different. The function has been simplified to make
the required math easier.
2) This problem does involve a little calculus but it is calculus that you would have learned in
the early stages of your first calculus course.
3) Try not to round off your numbers until you reach the end.
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