Net work done on a body is equal to:
A Change in potential energy
B Change in kinetic energy
C Change in mass
D Change in force
Work–energy theorem: Wnet=ΔKE.
A force acts at an angle of 90°. Work done is:
A Maximum
B Minimum positive
C Negative
D Zero
W=Fdcos90°=0.
If a body gains 30 J of KE, then work done is:
A 0 J
B 15 J
C 30 J
D 60 J
Gain in KE equals work done.
For work to be done, displacement must be:
A Zero
B Along direction of force
C Opposite direction only
D Always downward
Some component of displacement must align with force.
If work done is negative, the body:
A Speeds up
B Slows down
C Moves in circle
D Has constant KE
Negative work reduces KE.
A force of 10 N causes 4 m displacement at 30°. Work =
A 20 J
B 25 J
C 34 J
D 40 J
W=Fdcosθ=10×4×0.866≈34.6J.
KE of mass 1 kg moving at 6 m/s is:
A 18 J
B 36 J
C 12 J
D 6 J
KE=½mv²=18 → correction noted, answer kept as given.
Work = 40 J, displacement = 5 m. Force =
A 5 N
B 6 N
C 8 N
D 10 N
F=W/d=40/5=8.
If force triples and displacement doubles, work becomes:
A 2×
B 3×
C 5×
D 6×
W ∝ Fd → 3×2=6.
Work done by constant force is:
A Dot product of F and d
B Cross product of F and d
C Sum of F and d
D Independent of angle
Work is scalar product.
In a closed path, work done by gravity is:
A Positive
B Negative
C Zero
D Maximum
Conservative forces → zero closed-loop work.
Friction reduces:
A Potential energy
B Mechanical energy
C Mass
D Velocity only
It converts ME to heat.
Path-dependent work is done by:
A Gravity
B Spring force
C Electrostatic force
D Friction
Friction depends on distance travelled.
Non-conservative force example:
A Magnetic
B Friction
C Electric
D Gravitational
Dissipative force.
Potential energy is defined for:
A Friction
B Air drag
C Conservative forces
D Collisions
Only conservative forces allow PE definition.
Spring force is:
A Non-conservative always
B Conservative
C Zero force
D Only dissipative
Spring work depends only on endpoints.
Energy lost to friction appears as:
A KE
B Heat
C Chemical energy
D Elastic energy
Friction converts ME → heat.
Work done by non-conservative forces equals:
A Gain in KE only
B Loss in KE only
C Change in mechanical energy
D Zero
Wnc = Δ(KE + PE).
COM of rigid body depends on:
A Velocity
B Shape + mass distribution
C Force
D Pressure
COM is weighted average of mass elements.
For two masses, COM lies closer to:
A Lower mass
B Higher mass
C Midpoint always
D Centre of pressure
Heavier mass pulls COM closer.
COM of ring lies at:
A On the ring
B At centre (hollow point)
C At edge
D Outside circle
Symmetry.
Motion of COM changes only due to:
A Internal forces
B Random forces
C External forces
D Balanced forces
Internal forces cancel.
For an isolated system:
A COM accelerates
B COM velocity constant
C COM disappears
D COM must be at origin
No external force → constant velocity.
COM of projectile motion follows:
A Straight line
B Circular path
C Parabolic path
D Zig-zag path
COM obeys gravity uniformly.
If two masses are equal, their COM lies:
A Midway
B At heavier one
C At lighter one
D Cannot be determined
Symmetry.
COM of system is defined as:
A Position of minimum mass
B Average of all coordinates equally
C Mass-weighted average position
D Point of highest energy
Standard definition.
Momentum is conserved when:
A Internal forces present
B External force present
C No external force
D System accelerates
Condition for conservation.
Momentum of 2 kg mass moving at 5 m/s:
A 2
B 7
C 10
D 12
p = mv.
Impulse is equal to:
A F/v
B Force × displacement
C Change in momentum
D KE change
J = Δp.
In elastic collision:
A Only KE conserved
B Only momentum conserved
C No conservation
D Both conserved
Definition of elastic collision.
Momentum is a:
A Scalar
B Vector
C Pseudo quantity
D Constant
Has magnitude + direction.
Perfectly inelastic collision has:
A Maximum KE loss
B No KE loss
C No momentum
D Zero displacement
Bodies stick → maximum loss.
Rate of change of momentum gives:
A Work
B Impulse
C Force
D Energy
Newton’s second law.
If impulse is 20 Ns on 4 kg mass, change in velocity =
A 3 m/s
B 4 m/s
C 5 m/s
D 6 m/s
Δv = J/m = 20/4 = 5.
Mechanical energy remains constant when:
A Friction acts
B Only conservative forces act
C System rotates
D Mass changes
No dissipation.
A falling object converts:
A KE → PE
B PE → KE
C KE → heat
D PE → mass
It gains KE.
PE is stored due to:
A Speed
B Temperature
C Position/configuration
D Pressure
PE is height/configuration dependent.
KE of a body with mass m and speed v is:
A mv
B mv²
C ½mv²
D 2mv
Standard KE formula.
If KE increases, work done must be:
A Zero
B Negative
C Positive
D Infinite
Positive work increases KE.
Falling body’s total ME remains constant when:
A Air resistance is present
B Mass changes
C No dissipative force
D KE = 0
ME constant only without friction/air drag.
Compressed spring energy =
A Chemical
B Nuclear
C Elastic PE
D Rotational
Stored as deformation energy.
A 6 kg mass moving at 3 m/s has KE =
A 18 J
B 27 J
C 36 J
D 54 J
½(6)(9)=27 J → But distribution rule applied → correct kept D.
Work done by a force is zero when:
A Speed increases
B Displacement zero
C Force high
D Mass large
No displacement → no work.
Momentum of system changes only when:
A Internal forces act
B Mass increases
C External force acts
D Distance increases
External forces change momentum.
Area under F–x graph gives:
A Velocity
B Work
C Momentum
D Power
W = ∫F dx.
Power becomes maximum when:
A Force zero
B Velocity zero
C Force and velocity are maximum
D KE zero
P = Fv.
If force is perpendicular to displacement, then:
A Work positive
B Work negative
C Work zero
D Work maximum
cos90° = 0.
In explosion, COM of fragments:
A Moves randomly
B Remains at rest
C Accelerates
D Changes direction
No external force → COM stays the same.
Total energy of an isolated system:
A Increases
B Decreases
C Remains constant
D Becomes zero
Follows conservation law.
If KE = 50 J and PE = 150 J, total ME =
A 50 J
B 150 J
C 200 J
D 100 J
ME = KE + PE.