Unit 3: Work, Energy, and Power
Weightage : 15-25%
A small mass is released from rest at a very great distance from a larger stationary mass. Which of the following graphs best represents the gravitational potential energy \(U\) of the system of the two masses as a function of time \(t\)?
Correct Answer: \( \boxed{\mathrm{D}} \)
The gravitational potential energy of two masses is
\(U=-\dfrac{Gm_1m_2}{r}\)
where \(r\) is the separation between the masses.
Initially, the small mass is very far away, so \(r\rightarrow\infty\) and
\(U\rightarrow0\)
As the mass falls toward the larger mass, \(r\) decreases and the gravitational potential energy becomes increasingly negative.
Since the gravitational force increases as
\(F=\dfrac{Gm_1m_2}{r^2}\),
the object accelerates as it falls. Consequently, the potential energy decreases at an increasing rate, so the graph starts with a nearly zero slope and becomes progressively steeper in the negative direction.
This behavior is represented by Graph D.
Therefore, the correct answer is (D).
A mass \(m\) is attached to a mass \(3m\) by a rigid bar of negligible mass and length \(L\). Initially, the smaller mass is located directly above the larger mass, as shown above.
How much work is necessary to flip the rod \(180^\circ\) so that the larger mass is directly above the smaller mass?
Correct Answer: \( \boxed{\mathrm{B}} \)
The work required equals the increase in the system’s gravitational potential energy.
During the \(180^\circ\) rotation:
• The larger mass \(3m\) is raised by a vertical distance \(L\), increasing its potential energy by
\(\Delta U_{3m}=3mgL\)
• The smaller mass \(m\) is lowered by a vertical distance \(L\), decreasing its potential energy by
\(\Delta U_{m}=-mgL\)
Therefore, the total change in gravitational potential energy is
\(\Delta U=3mgL-mgL=2mgL\)
Since the rod is moved slowly, the external work required is equal to the increase in gravitational potential energy:
\(W_{\mathrm{ext}}=\Delta U=2mgL\)
Therefore, the correct answer is (B).
A particle is subjected to a conservative force whose potential energy function is
\(U(x)=(x-2)^3-12x\)
where \(U\) is given in joules when \(x\) is measured in meters.
Which of the following represents a position of stable equilibrium?
Correct Answer: \( \boxed{\mathrm{E}} \)
Equilibrium occurs where the force is zero. Since
\(F(x)=-\dfrac{dU}{dx}\),
the equilibrium positions satisfy
\(\dfrac{dU}{dx}=3(x-2)^2-12=0\)
\((x-2)^2=4\)
\(x=0\) or \(x=4\)
To determine whether an equilibrium point is stable, evaluate the second derivative:
\(\dfrac{d^2U}{dx^2}=6(x-2)\)
At \(x=0\):
\(\dfrac{d^2U}{dx^2}=6(0-2)=-12<0\),
so \(x=0\) is a point of unstable equilibrium (a local maximum of \(U\)).
At \(x=4\):
\(\dfrac{d^2U}{dx^2}=6(4-2)=12>0\),
so \(x=4\) is a point of stable equilibrium (a local minimum of \(U\)).
Therefore, the correct answer is (E).
