Determining Potential Energy Values — Question 1

Question 1

Choosing a gravitational datum below the laboratory origin

A 3 kg object has upward coordinate y measured from a laboratory floor. Gravity is uniform with $g=10$ m/s$^2$. Define gravitational potential energy to be zero at y=$-2$ m.

1. Determine U(y).

2. Find U at y=1 m and the change from y=1 to y=4 m.

3. What changes if the laboratory floor is chosen as the zero instead?

Question 1 - Solution

Choosing a gravitational datum below the laboratory origin

(a) Apply the reference through a definite integral.

Use the specified reference in the definite integral: $$U(y)-U(-2)=-\int_{-2}^{y}(-mg)\,ds=30(y+2).$$ $$\boxed{U(y)=30(y+2)\ \mathrm J}\quad(y\text{ in m}).$$

(b) Evaluate the function and the difference.

Evaluate both states with the same datum: $$\boxed{U(1)=90\ \mathrm J},\qquad U(4)=180\ \mathrm J,\qquad\boxed{\Delta U=180-90=90\ \mathrm J}.$$

(c) Shift only the additive constant.

Choosing the floor as zero subtracts a constant: $$\widetilde U(y)=30y=U(y)-60\ \mathrm J,\qquad\Delta\widetilde U=\Delta U=90\ \mathrm J.$$ Forces and gravitational works are unchanged.