The figure shows a thin rod, of length L = 1.4 m and negligible mass, that can p

Question

The figure shows a thin rod, of length L = 1.4 m and negligible mass, that can pivot about one end to rotate in a vertical circle. A heavy ball of mass m = 6.2 kg is attached to the other end. The rod is pulled aside to angle theta_0 = 14 degree and released with initial velocity v_0 = 0. As the ball descends to its lowest point, how much work does the gravitational force do on it and what is the change in the gravitational potential energy of the ball-Earth system? If the gravitational potential energy is taken to be zero at the lowest point, what is its value just as the ball is released?

Explanation / Answer

given that

L = 1.4 m

m = 6.2 kg

theta = 14 degree

v0 = 0 m/s

part(a)

we know that

change in rod potential energy

delta PE = m*g*delta H

where deltaH = L - L*cos(theta)

deltaH = 1.4 - 1.4*cos(14)

deltaH = 0.0415 m

delta PE = 6.2*9.81*0.0415

delta PE = 2.52 J

work done by gravitational force is 2.52 j.

part(b)

change in gravitational potential energy = -W

change in gravitational potential energy = -2.52 J

part(c)

final PE -initial PE = -2.52 J

0 - initial PE = -2.52 J (given final potential energy = 0)

so

initial PE = 2.52 J

answer

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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