A particle of mass m is displaced through a small vertical distance y near the E

Question

A particle of mass m is displaced through a small vertical distance y near the Earth's surface. Show that in this situation the general expression for the change in gravitational potential energy given by the equation reduces to the familiar relationship ?U = mg ?y

A student is studying the potential energy change of a 50 kg object raised 100 km above Earth's surface. What will be the percentage error if he simply used the approximate relation ?U = mg?y?
percent error =

%
Write expressions for the potential energy change as described in the question, and evaluate the percentage difference between them.

Explanation / Answer

Let me show you the normal way to show exponents here. 23.12 m/s^2 where ^ is shift+6. particle : Think about an oscillating spring. It goes back and forth across the equilibrium position. When off to either side of the equilibrium position, the spring tries to pull it toward the equilibrium position. In other words, if it's off to the left, the spring pulls it right. The further away it is, the harder the pull is. pendulum: The formula is T = 2*pi*sqrt(l/g) where l = 1 m and g = 23.12 m/s^2 SHM: 3 of these are correct. They should sound at least probable.But it's probably easiest to pick out the one that sounds really wrong. Think about a pendulum again -- it swings off to the side, stops, and goes back the other way until it stops off to the other side.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.