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Work, Energy And Power Mcq
Quiz 7
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Q.1
A spring of natural length l and spring constant k is fixed on the ground and the other is fitted with a smooth ring of mass m which slides on a horizontal rod fixed at heigh also equal to l as shown in figure. Initially the spring makes an angle 37° with the vertical when the system is released from rest. What is the speed of the ring when the spring becomes vertical?
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a)
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b)
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c)
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d)
Explanation
From the initial position of the ring as shown in figure The length of spring=l / cos37=5l/4 ∴ Extension=5l/4 - l=l/4 Energy stored in spring E This stored energy when released becomes kinetic energy of the ring. If v is the velocity of the ring kinetic energy when it is vertical==½ mv2 By the principle of conservation of energy Answer: (d)
Q.2
A particle is whirled 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
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a)√gl
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b) √2gl
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c)√3gl
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d)√5gl
Explanation
Let minimum speed is v at highest point (A) let U is the minimum speed at point B At this position centrifugal force is upward and gravitational force is down ward thus mv2 / l=mg v2=gl --eq(1) Using the principle of conservation of energy between point A and B ½ mU2=mgl + ½ mv2 substituting value of v2 from equation (1) in above equation we get ½ mU2=mgl + ½ mgl=(3/2)mgl U=√(3gl)Answer: (c)
Q.3
A body is slowly lowered on a massive platform moving horizontally at a speed of 4m/s , through what distance will the body slide relative to the platform? [ The coefficient of friction is 0.2; g=10 m/s2]
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a) 4 m
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b) 16 m
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c)4 m
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d)20 m
Explanation
The frictional force between the body and the platform=µmg, where m is the mass of the body Initially the relative velocity=4 m/s The relative retardation=µg=0.2 × 10=2 m/s2 If S id the relative displacement before the relative velocity becomes zero, we have for formula v2=u2 + 2as0=42 - 2 × 2 × S S=16/4=4 m Answer: (c)
Q.4
A child builds a tower from three blocks. The blocks are uniform cubes of side 2cm. The block are initially lying on the same horizontal surface and each block has a mass of 0.1kg. The work done by the child is
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a) 4 J
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b) 0.04 J
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c)6 J
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d)0.06 J
Explanation
Work done by the child=Increase in potential energy of the block Ground block=0 First block work=(0.1g)(2 × 10-2) Second block work=(0.1g)(4 × 10-2) total work=0.06J Answer:(d)
Q.5
A block slides down an inclined plane of slope θ with constant velocity . It is then projected up the plane with an initial velocity u. How far up the incline will it move before coming to rest.
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a) u2/ gsinθ
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b) u2/ 2gsinθ
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c) u2/ 4gsinθ
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d) 2u2/ gsinθ
Explanation
Constant velocity indicate frictional force acting against gravitational force thus mgsinθ=µmgcosθ For upward motion retardation is=downward gravitional force + frictional force=2gsinθ, now ½ mu2=2mgsinθ × l l=u2/ 4gsinθ Answer: (c)
Q.6
A mass of 1 kg suspended on a thread deviates through an angle 30°. Find the tension of the thread at the moment the weight passes through the position of equilibrium
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a)12.4 N
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b) 15 N
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c)24.8 N
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d)6.2 N
Explanation
At the moment the weight passes through the position of equilibrium the tension of the thread By the conservation of energy mgh=½ m v2v=√2ghBut h=l(1 - cos30) substituting value of mv2 /l in equation for tension we getT=mg [ 1 + 2(1 - cos30)] Given m=1 kg ; g=9.8 m/s2; cos30=√3/2 ∴ T=12.4 N Answer: (a)
Q.7
A block of mass m moving with speed v compresses a spring through a distance x before its speed is halved. What is the value of spring constant
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a)
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b)
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c)
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d)
Explanation
Initial kinetic energy=½ mv2 Final energy=½ m (v/2)2 + ½ k x2 By principle of conservation of energy Answer: (d)
Q.8
The work done against gravity in moving the block of mass m through a distance S up the slope is
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a) mh
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b) mgS
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c)mS
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d)mgh
Explanation
Answer:(d)
Q.9
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 friction on the block in time t will be
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a) zero
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b) mgvtcos2θ
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c) mgvtsin2θ
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d) mgvtsinsin2θ
Explanation
Since Block is at rest frictional force f=mgsinθ dispacement=vtsinθ Work W=f.S=mgsinθ.vt.sinθ=mgvtsin2θ Answer: (c)
Q.10
A car is moving with constant speed of 20 m/s against a resistance of 100N. The power exerted by the car is
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a)2 kW
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b) 5 W
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c)200 W
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d)1 kW
Explanation
P=F.vAnswer: (a)
Q.11
Two equal masses are attached to the two ends of the 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 masses is
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a) ½ kx2
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b) - ½ kx2
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c)¼ kx2
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d)- ¼ kx2
Explanation
Energy stored in spring=½ kx2 Total work done by spring=-½ mv2 work done on each mass=by the string=- ¼kx2Answer: (d)
Q.12
The work done by the external force on a system equals the change in
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a) total energy
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b) kinetic energy
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c)potential energy
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d)none of these
Explanation
Answer:(a)
Q.13
Two springs of spring constants 1000 N/m and 2000 N/m are stretched with same source. They will have potential energy in the ratio of
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a) 2:1
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b) 22 : 12
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c) 1 : 2
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d) 12 : 22
Explanation
Potential energy of spring=½ k x2 and F=kx thus x=F/k Potential energy=½ k ( F/k)2 Potential energy=F2/2k F2=E × 2k Given force on both spring is same thus E1 × 2k1=E2 × 2k2 E1 / E2=k2/k1=2000/1000=2:1 Answer: (a)
Q.14
A ball falls under gravity from height 10 m with an initial velocity u. It hits the ground, loses 50% of the energy in collision and it rises to the same height. What is the value of u
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a)14 m/s
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b) 7 m/s
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c)28 m/s
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d)9.8 m/s
Explanation
Let v be the velocity when it hits the ground.Then v2 - u2=2asv2 - u2=2 × 9.8 × 10v2=u2 + 196Now let v' be the velocity after imapact and it reaches the same height 10 m V'2 - 0=2 × 9.8 × 10V'2=196V'=14 m/s Given there is 50% energy loses Ratio of kinetic energy loses thus before impact and after impact=2 Answer: (a)
Q.15
The potential energy of a particle varies with position x according to the relation U(x)=x3 - 4x . The point x=2 is point of
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a) stable equilibrium
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b) Unstable equilibrium
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c)neutral equilibrium
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d)none of the above
Explanation
F=-dU/dx=-3x2 + 4 at x=2 , F=-8 unit and U=0 Since, force is not zero, hence it is not in equilibriumAnswer: (d)
Q.16
Consider two observers moving with respect to each other at a speed v along a straight line. They observe a block of mass m moving a distance l on a rough surface. The following quantities will be same as observed by the two observer
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a) kinetic energy of the block at time t
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b) work done by friction
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c)total work done on the block
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d)acceleration of block
Explanation
Since both observers have same and uniform velocity they are in inertial frame., the acceleration of the block will be same Answer:(d)
Q.17
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
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a) vertically downward
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b) vertically upward
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c) horizontally
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d) the speed does not depend on the initial direction
Explanation
Answer: (d)
Q.18
An ideal massless spring S can be compressed one metre by a force of 100N. The same spring is placed at the bottom of a frictionless inclined plane inclination at 30° to horizontal. A block M of mass 10 kg is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring 2 metre. What is the speed of the mass just before it reaches the spring
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a)√20 m/s
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b) √30 m/s
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c)√10 m/s
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d)√40 m/s
Explanation
Applied force on the spring, F=kxx=F/k=100/1=100 N/m Let the mass M slides a distance S metres along the inclination before hitting the spring. The spring gets compressed by 2 metres. Hence the mass M slides a total distance ( S + 2) metre along the incline . From the geometry of figure height of M=(S + 2)sin30 above the bottom . Thus h=(S + 2)/2 Potential energy of mass M=Mgh=Mg(S + 2)/2 When the spring is compressed, energy is converted in potential energy of spring Mass M has falls at height of Ssin30=S/2=1 m before colliding with spring Gain in K.E=Loss in P.E ½ M v2=Mgh v=√(2gh) v=√( 2 × 10 × 1)=√20 m/sAnswer: (a)
Q.19
A small block of mass 100g is pressed against a horizontal spring fixed at one end and compression is 5cm. The spring constant is 100 N/m. When the block moves horizontally it leaves the spring. Where will it hit the ground 2m below the spring
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a) 1.5 m from free end of the spring
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b) Horizontal distance of 2 m from end of the spring
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c)0.5 m from free end of the spring
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d)Horizontal distance of 1 m from end of the spring
Explanation
Let v be the velocity when it leaves the spring Then kinetic energy of block=potential energy of spring ½ mv2=½ kx2 Time to fall vertical distance of 2 metres from spring Horizontal distance d=velocity × t Answer: (d)
Q.20
The work done by all the forces ( external and internal) on a system equals the change in
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a) total energy
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b) kinetic energy
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c)potential energy
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d)non of these
Explanation
Answer:(b)
Q.21
A body is projected at an angle of 30° to the horizontal with kinetic energy 40 J. What will be the kinetic energy at the top-most point?
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a) 25 J
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b) 40 J
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c) 30 J
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d) 20 J
Explanation
At the topmost point , the horizontal component is ucosθ Initial kinetic energy=½ mu2=40 J Kinetic energy at topmost point=½ m u2 cos2 θ on substituting value of ½mu2 we get Kinetic energy at topmost point=40cos2θ Kinetic energy at topmost point=40 × 3 / 4=30 J Answer: (c)
Q.22
A bob is suspended from a crane by a cable of length l=5m. The crane and load are moving at a constant speed v. the crane is stopped by a bumper and the bob on the cable swings out an angle of 60°. The initial speed v is ( g=9.8 m/s2)
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a)10 m/s
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b) 7 m/s
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c)4 m/s
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d)2 m/s
Explanation
Change in kinetic energy=potential energy of bob ½ mv2=mgl(1 - cos60) Answer: (b)
Q.23
A block of mass 1kg sliding down a curved track that is one quadrant of a circle of radius 1m. Its speed at the bottom is 2m/s. The work done by the frictional force is
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a) -8J
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b) +8J
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c)9J
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d)-9J
Explanation
Here work done by frictional force=change in kinetic energy - loss in potential energy work done by frictional force=( ½ × 1 × 4) - ( 1 × 10 × 1)=-8J Answer: (a)
Q.24
When a spring is stretched by 2 cm, it stores 100 J of energy. If it is stretched further by 2 cm, the stored energy will be increased by . [Orissa JEE 2002]
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a)100 J
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b) 300 J
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c)200 J
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d)400 J
Explanation
Given when x=2 cm=2 × 10-2 , enrgy U=100J Thus from formula for potential energy of spring 100=½ k × (2 × 10-2) 2 Thus k=50 × 104 Now change in potential energy=½ k ( x'2 - x2) By substituting value of x'=4 cm=4 × 10-2 and x=2 cm=2 × 10-2 and We get ΔU=300J Answer:(b)
Q.25
A spring with spring constant k when stretched through 1 cm, the potential energy is U. If it is stretched by 4 cm. The potential energy will be . [Orissa PMT 2004]
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a) 4U
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b) 16 U
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c) 8U
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d) 2U
Explanation
Potential energy U=(1/2)kx2 so U is propotinal to x2 that is why If elongation made 4 times then potential energy will become 16 times. Answer: (b)
Q.26
A weightless rigid rod AB of length l carries two equal masses m , one secured at the end and other at the middle of the rod as shown in figure. The rod can rotate in vertical plane around the hinge at A. The minimum horizontal velocity required to be imparted to the end B of rod so as to make the rod go around in a complete circle is
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a)√(4gl)
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b) √(5gl)
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c)√(24gl/5)
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d)√(24gl/7)
Explanation
If v be the velocity of B, then v/2 is the velocity of C.From conservation of energy Answer: (c)
Q.27
The potential energy of a particle varies with position x according to the relation U(x)=x3 - 4x . The point x=2 is point of
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a) stable equilibrium
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b)unstable equilibrium
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c)neutral equilibrium
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d)none of the above
Explanation
F=-dU/dx=-3x2 + 4 , at x=2, F=-8 unit and U=0 Since force is not zero, hence it is not in equilibriumAnswer: (d)
Q.28
The kinetic energy acquired by a body of mass m is travelling some distance s, starting from rest under the actions of a constant force, is directly proportional to.. [Pb. PET 2000]
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a)mo
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b)m2
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c)m
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d) √m
Explanation
From equation of motionv2=u2 + 2aSsince u=0 v2=2aS but a=F/m v2=2(F/m)S Now kinetic energy=½ mv2 Kinetic energy=½ × m × [ 2(F/m)S] Thus kinetic energy ∝ mo Answer:(a)
Q.29
A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to . [AIEEE 2004]
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a) x2
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b) x
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c) ex
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d) logex
Explanation
This condition is applicable for simple harmonic motion. As particle moves from mean position to extreme position its potential energy increases according to expression U=(1/2)kx2 and accordingly kinetic energy decreases Answer: (a)
Q.30
A body at rest may have
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a)Energy
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b)Speed
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c)Momentum
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d)Velocity
Explanation
Answer: (a)
0 h : 0 m : 1 s
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