Work, Energy, Power and Conservation Laws: questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
A force F=k(yi^+2xj^) acts on a particle moving in the xy-plane, where k is a positive constant and x, y are in metres. The particle is taken from the origin O(0,0) to the point P(a,a) (a>0) along two different paths: path 1 is the straight line y=x; path 2 is the parabola y=x2/a. Let W1 and W2 be the work done by F along paths 1 and 2 respectively. The value of W2−W1 is ______. Exactly one option is correct.
A0, because the work depends only on the end points
B−6ka2
C6ka2
D3ka2
Show answer and solution
Answer: C
Path 1 (y=x, dy=dx): W1=∫0ak(x+2x)dx=23ka2.
Path 2 (y=x2/a, dy=2xdx/a): W2=∫0ak(ax2+2x⋅a2x)dx=∫0aa5kx2dx=35ka2.
So W2−W1=6ka2.
Check: ∂Fy/∂x−∂Fx/∂y=2k−k=k=0, so the force is non-conservative and the loop integral (out along the parabola, back along the line, i.e. anticlockwise) equals k×(area between the curves) =k(2a2−3a2)=6ka2.
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
A uniform flexible chain of length L lies straight on a rough horizontal table, perpendicular to the table's edge, with a length L/2 hanging vertically over the edge. The coefficient of friction (static and kinetic) between the chain and the table is μ=1/2; the edge is smooth and its curvature is negligible. The chain is released from rest and slides off without bunching. Acceleration due to gravity is g. The speed of the chain at the instant its last link leaves the table is ______. Exactly one option is correct.
A85gL
B43gL
C2gL
D87gL
Show answer and solution
Answer: A
It starts: hanging weight λgL/2 exceeds maximum friction μλgL/2=λgL/4 (λ = mass per length).
Let x be the hanging length. Friction is μλg(L−x), a force that changes as the chain moves, so its work is Wf=−∫L/2Lμλg(L−x)dx=−2μλg(2L)2=−16λgL2.
Gravity (reference: table level): initial PE =−λ2Lg4L=−8λgL2; final PE =−λLg2L=−2λgL2; so Wg=83λgL2.
Work-energy: 21λLv2=83λgL2−161λgL2=165λgL2, so v=5gL/8.
(Ignoring friction gives 3gL/4; treating friction as constant μλgL/2 over the whole L/2 gives gL/2.)
Difficulty: Very hard
Question 3Level 3 · AdvancedSingle correct
A wedge of mass M with a smooth inclined face of inclination θ rests on a smooth horizontal floor. A small block of mass m is placed on the inclined face and the system is released from rest. The block slides down the face without losing contact, and the wedge slides along the floor without tipping. Acceleration due to gravity is g. When the block has descended a vertical height h (it is still on the face), the speed of the wedge is ______. Exactly one option is correct.
A(M+m)(M+mcos2θ)2m2ghcos2θ
BmcosθM(M+m)2gh
C(M+m)(M+msin2θ)2m2gh
D(M+m)(M+msin2θ)2m2ghcos2θ
Show answer and solution
Answer: D
Let the wedge move with speed V (backwards) and the block with speed u relative to the wedge along the face.
Horizontal momentum (no horizontal external force): m(ucosθ−V)=MV⇒ucosθ=m(M+m)V.
Block's ground velocity: horizontal ucosθ−V=mMV, vertical usinθ.
Energy (normal forces do zero total work, since the contact is smooth and the constraint is workless for the pair): mgh=21MV2+21m[m2M2V2+u2sin2θ].
Substituting u=mcosθ(M+m)V: mgh=2V2[mM(M+m)+m(M+m)2tan2θ]=2mcos2θ(M+m)V2(M+msin2θ).
Hence V=(M+m)(M+msin2θ)2m2ghcos2θ.
Limits: M→∞ gives V→0; θ→90∘ gives V→0, as expected.
Difficulty: Very hard
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