Question

Consider a uniform gravitational field (a fair approximation near the surface of a planet). Find U(yf)−U(y0)=−∫yfy0F⃗ g⋅ds⃗ , where F⃗ g=−mgj^ and ds⃗ =dyj^.

Answer 1

Answer:

U(yf) − U(y0) = mg( yo - yf )

Step-by-step explanation:

We have that

$U(y_f)-U(y_o)=-\int_{y_f}^{y_o}\textbf{F}_g\cdot d\textbf{s}\\\textbf{F}_g=mg\hat{j}\\d\testbf{s}=dy\hat{j}$

Hence we have

$F_g \cdot ds =-mgdy$

and by replacing in the integral we have

$-\int_{y_f}^{y_o}(-mgdy)=mg[y_o-y_f]$

hope this helps!

Answer 2

The difference in gravitational potential work is $U(y_{f}) - U(y_{o}) = m\cdot g\cdot (y_{f}-y_{o})$.

A uniform gravitational field is a good approximation when it is given that distance by the object is small in comparison with the dimensions of the planet and it is near to the surface of the planet. Gravitational acceleration can be considered approximately constant:

$g \approx \frac{G\cdot M}{r^{2}}$ (1)

Where:

• $G$ - Gravitational constant, in cubic meters per kilogram-square second.
• $M$ - Mass of the planet, in kilograms.
• $r$ - Radius of the planet, in meters.

Then, we can reduce the work-energy equivalence by means of this formula:

$dU = \vec F\,\bullet\,d\vec s$ (2)

$dU = \|\vec F\|\cdot \|d\vec s\|\cdot \cos \theta$

Since both gravitational force and distance are parallel to each other, then we reduce the expression into this:

$dU = F\cdot ds$

$U = \int\limits^{y_{f}}_{y_{o}} {F} \, dy$

$U(y_{f}) - U(y_{o}) = F\cdot (y_{f}-y_{o})$

$U(y_{f}) - U(y_{o}) = m\cdot g\cdot (y_{f}-y_{o})$

The difference in gravitational potential work is $U(y_{f}) - U(y_{o}) = m\cdot g\cdot (y_{f}-y_{o})$.

We kindly invite to see this question on gravitational potential energy: https://brainly.com/question/8822715