Conservative and Non Conservative Forces
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Give three examples when work done is zero.
- Force acting on the system will be (B−2Ax)
- At equilibrium, potential energy will be −B24A
- system is in stable equilibrium.
- None
Where α and β are dimensional constants.
The angular momentum of the particle becomes the same, as it was for t=0, at time
- (20yz)^i+(20xz)^j+(20xyz2)^k
- −(20yz)^i−(20xz)^j+(20xyz2)^k
- −(20yz)^i−(20xz)^j−(20xyz2)^k
- (20yz)^i+(20xz)^j−(20xyz2)^k
A particle free to move along the x-axis has potential energy given by U(x)=k[1−exp(−x)2] for −∞≤x≤+∞, where k is a positive constant of appropriate dimensions. Then
If its total mechanical energy is k/2, it has its minimum kinetic energy at the origin
For any finite non-zero value of x, there is a force directed away from the origin
For small displacements from x = 0, the motion is simple harmonic
At point away from the origin, the particle is in unstable equilibrium
- 7 J
- 3 J
- 5 J
- zero
A body is moving up an inclined plane of angle θ with an initial kinetic energy E. The coefficient of friction between the plane and the body is μ. The work done against friction before the body comes to rest is
μ cos θE cos θ+sin θ
μE cos θ
μE cos θμ cos θ−sin θ
μE cos θμ cos θ+sin θ
Work done in the motion of a body over a closed loop is zero for every force in nature.
- True
- False
- 2√Gd(M1+M2)
- 2√2Gd(M1+M2)
- 2√Gmd(M1+M2)
- 2√Gm(M1+M2)d(R1+R2)
- →F=−3y^i−4x^j
- →F=−5y^i−5x^j
- →F=3y^i+4x^j
- →F=5y^i−5x^j
Work done by conservative forces is equal to increase in potential energy.
- True
- False
- 10 m
- 12 m
- 15 m
- 18 m
A particle of mass m moves with constant speed along a circular path of radius r under the action of a force F. Its speed is:
√rFm
√Fr
√Fmr
√Fmr
- Total energy remains constant
- Work done in round trip is zero
- Work done is dependent on path
- Work done is completely recoverable
List - I | List - II |
P. →r(t)=αt^i+βt^j | 1. →p |
Q. →r(t)=αcos(ωt)^i+βsin(ωt)^j | 2. →L |
R. →r(t)=α(cos(ωt)^i+sin(ωt)^j) | 3. K |
S. →r(t)=αt^i+β2t2^j | 4. U |
5. E |
- P→1, 2, 3, 4, 5; Q→2, 5; \ R \rightarrow 2, 3, 4, 5 ; \ S \rightarrow 5$
- P→1, 2, 3, 4, 5; Q→3, 5; \ R \rightarrow 2, 3, 4, 5 ; \ S \rightarrow 2, $
- P→2, 3, 4; Q→5;\ R \rightarrow 1, 2, 4 ; \ S \rightarrow 2, 5$
- P→1, 2, 3, 5; Q→2, 5; \ R \rightarrow 2, 3, 4, 5 ; \ S \rightarrow 2, 5$
- Minimum speed V0 to slide full length of the track is √60m/s
- Minimum speed V0 to reach D is √105m/s
- Normal force on the object by the track at C if it reaches D is 15 N
- Normal force on the object by the track at B if it reaches D is 32.5 N
Work done by a conservative force on a system is equal to
The change in kinetic energy of the system
The change in potential energy of the system
The change in total mechanical energy of the system
None of the above
- The work done by air resistance increases if the ball bounces more number of times.
- The kinetic energy of the ball increases if the ball bounces more number of times.
- The potential energy of the ball increases as the ball bounces.
- The work done by gravity increases if the ball bounces more number of times.
- 8 N
- 10 N
- 6 N
- 4 N
- 60 J
- 32 J
- 68 J
- 36 J
- False
- True
- U1=U2
- U1>U2
- U1<U2
- U1≥U2
where a and b are positive constants. Find the equilibrium position and discuss whether the equilibrium is stable, unstable or neutral.
- x=−b2a is the stable equilibrium position.
- x=0 is the stable equilibrium position
- x=b2a is a unstable equilibrium position
- x=b2a is the stable equilibrium position
- False
- True
- →F must be conservative
- →F may be conservative
- →F must be non-conservative
- →F may be non-conservative
- 7 J
- 3 J
- 5 J
- zero