A pendulum bob with a mass of 0.20 kg is attached to a string with a length of 0

Question

A pendulum bob with a mass of 0.20 kg is attached to a string with a length of 0.81 m. We choose the potential energy to be zero when the string makes an angle of 90^circ with the vertical.

a.Find the potential energy of this system when the string makes an angle of 45^circ with the vertical.
b. Is the magnitude of the change in potential energy from an angle of 90^circ to 45^circ greater than, less than, or the same as the magnitude of the change from 45^circ to 0^circ? Explain.
c.Calculate the potential energy of the system when the string is vertical.

Explanation / Answer

a) h = -0.81*sin(45)

g= 9.81

m = 0.2

PE1 = m*g*h

PE1 = -1.124 J

c) h = -0.81

PE2 = m*g*h

PE2 = -1.589 J

b) change during 90 to 45 degree = dPE1 = -1.124J

change during 45 to 0 degree = dPE2 = -1.589+1.124 = -0.465

ignoring the minus sign while seeing the magnitude,

as dPE1>dPE2,

he magnitude of the change in potential energy from an angle of 90^circ to 45^circ greater than the magnitude of the change from 45^circ to 0^circ

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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