Q#1
A heavy stone is thrown from a cliff of height h with a speed v. The stone will hit the ground with maximum speed if it is thrown(a) vertically downward
(b) vertically upward
(c) horizontally
(d) the speed does not depend on the initial direction
Answer: (d)
When the stone is thrown its total mechanical energy
= KE + PE
= ½mv² + mgh (i)
This energy is conserved and when near the ground its P.E. is also converted to K.E. because of h=0. suppose it hits the ground with speed v', its total energy = ½mv'² (ii)
equating (i) and (ii) we get,
½mv'² = ½mv² +mgh
v'² = v² + 2gh
Clearly v' depends only on the value of v whatever be the initial direction.
Q#2
Two springs A and B (kA = 2kB) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, that in B is
(a) E/2
(b) 2E
(c) E
(d) E/4
Answer: (b)
Since (kA = 2kB) it means for the same force xB = 2xA . So if energy stored in spring A = E = FxA
Energy stored in spring B = FxB = F(2xA) = 2E
Q#3
Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled out symmetrically to stretch the spring by a length x over its natural length. The work done by the spring on each mass is
(a) ½kx²
(b) -½kx²
(c) ¼kx²
(d) -¼kx²
Answer: (d)
The work done by the spring on both of the masses is equal to the potential energy stored in the spring. The potential energy stored in the spring = -½kx². Since the masses are pulled out symmetrically, the work done by the string on each mass
Q#4
The negative of the work done by the conservative internal forces on a system equals the change in
(a) total energy
(b) kinetic energy
(c) potential energy
(d) none of these
Answer: (c)
The Potential energy of a system is due to its configuration. When conservative internal forces do work on the system, the potential energy changes to kinetic energy. So negative of work done by the conservative internal forces will be the change in potential energy.
Q#5
The work done by the external forces on a system equals the change in
(a) total energy
(b) kinetic energy
(c) potential energy
(d) none of these
Answer: (a)
External forces will either change KE or PE or both of a system. Since total mechanical energy (KE + PE) of a system is conserved when no external force acts, hence work done by external force will change total mechanical energy.
Q#6
The work done by all the forces (external and internal) on a system equals the change in
(a) total energy
(b) kinetic energy
(c) potential energy
(d) none of these
Answer: (b)
Though work done by external forces will change the total energy of the system, work done by the internal forces will increase kinetic energy only. So, when both do work KE is sure to change.
Q#7
___________of a two-particle system depends only on the separation between the two particles. The most appropriate choice for the blank space in the above sentence is
(a) kinetic energy
(b) total mechanical energy
(c) potential energy
(d) total energy
Answer: (c)
The potential energy of a system is due to its configuration and separation depicts the configuration.
Q#8
A small block of mass m is kept on a rough inclined surface of inclination θ fixed in an elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be
(a) zero
(b) mgvt cos²θ
(c) mgvt sin²θ
(d) mgvt sin2θ
Answer: (c)
In time t, vertical distance covered = vt
The vertical component of frictional force = fsinθ
With f = µmgcosθ
Work done by this component
= fsinθ(vt)
= µmgcosθ sinθ(vt)
= (sinθ/cosθ) mg(vt)cosθsinθ (Since µ = tanθ)
= (mgvt) sin²θ
Q#9
A block of mass m slides down a smooth vertical circular track. During the motion, the block is in
(a) vertical equilibrium
(b) horizontal equilibrium
(c) radial equilibrium
(d) none of these
Answer: (d)
Self Explanatory.
Q#10
v² = gl
Now let the speed when string is horizontal = v'
Here total energy = ½mv'² = ½mv² + mgl
v'² = v² + 2gl = gl + 2gl = 3gl
v' = √(3gl)
The vertical component of frictional force = fsinθ
With f = µmgcosθ
Work done by this component
= fsinθ(vt)
= µmgcosθ sinθ(vt)
= (sinθ/cosθ) mg(vt)cosθsinθ (Since µ = tanθ)
= (mgvt) sin²θ
Q#9
A block of mass m slides down a smooth vertical circular track. During the motion, the block is in
(a) vertical equilibrium
(b) horizontal equilibrium
(c) radial equilibrium
(d) none of these
Answer: (d)
Self Explanatory.
Q#10
A particle is rotated in a vertical circle by connecting it to a string of length l and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is
(a) √(gl)
(b) √(2gl)
(c) √(3gl)
(d) √(5gl)
Answer: (c)
Let speed at the top = v, it is balanced by weight mg,
so, mv²/l = mg
(a) √(gl)
(b) √(2gl)
(c) √(3gl)
(d) √(5gl)
Answer: (c)
Let speed at the top = v, it is balanced by weight mg,
so, mv²/l = mg
v² = gl
Now let the speed when string is horizontal = v'
Here total energy = ½mv'² = ½mv² + mgl
v'² = v² + 2gl = gl + 2gl = 3gl
v' = √(3gl)
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