Q 1 :

A force (3x2+2x-5)N displaces a body from x=2m to x=4m. Work done by this force is __________ J.             [2024]



(58)       Force, F=(3x2+2x-5)N,x1=2m and x2=4m

              Work done, W=∫x1x2F·dx=∫24(3x2+2x-5)dx

               W=[x3]24+[x2]24-5[x]24=[64-8]+[16-4]-5[4-2]=58J

 



Q 2 :

A body of mass 50 kg is lifted to a height of 20 m from the ground in the two different ways as shown in the figures. The ratio of work done against the gravity in both the respective cases, will be         [2024]

  • 2 : 1

     

  • 1 : 2

     

  • 3 : 2

     

  • 1 : 1

     

(4)

Work done by gravity is path independent, it depends only on vertical displacement, so same work is done in both paths.

 



Q 3 :

A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is    [2024]

  • 1 m

     

  • 6 m

     

  • 5 m

     

  • 2 m

     

(4)

sin30°=h10⇒h=5

At maximum compression

using Work-Energy Theorem,

Wgravity+Wfriction+Wspring=ΔKE

mgh-μmg(2+x)-12kx2=0-0

5×10×5-12×5×10(2+x)-12100×x2=0

250-50x2-[50+25x]=0

200-50x2-25x=0

2x2+x-8=0

x=-1±12+4×2×82×2⇒x=-1±654≈1.77m

Nearest answer is 2m.



Q 4 :

A block of mass 100 kg slides over a distance of 10 m on a horizontal surface. If the coefficient of friction between the surfaces is 0.4, then the work done against friction (in J) is       [2024]

  • 4200

     

  • 3900

     

  • 4000

     

  • 4500

     

(3)

Given m=100kg, μ=0.4

Friction force, f=μmg=0.4×100×10=400N

Now W=f·s=400×10=4000J

 



Q 5 :

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60 by a force of 10 N parallel to the inclined surface as shown in the figure. When the block is pushed up by 10 m along the inclined surface, the work done against frictional force is [g=10m/s2]     [2024]

  • 53 J

     

  • 5 J

     

  • 5×103 J

     

  • 10 J

     

(2)

fk=μkN=μkmgcos60°

fk=0.1×10×12=0.5 N

Wfriction=-fk·S

                 =-(0.5)×10

Wfriction=-5 J



Q 6 :

A force F→=2i^+bj^+k^ is applied on a particle and it undergoes a displacement i^–2j^–k^. What will be the value of b, if work done on the particle is zero.          [2025]

  • 0

     

  • 12

     

  • 13

     

  • 2

     

(2)

Given, W = 0

∴ F→·S→=0

(2i^+bj^+k^)·(i^–2j^–k^)=0

2–2b–1=0 ⇒ b=12



Q 7 :

A force F=α+βx2 acts on an object in the x-direction. The work done by the force is 5 J when the object is displaced by 1 m. If the constant α = 1 N then β will be          [2025]

  • 15 N/m2

     

  • 10 N/m2

     

  • 12 N/m2

     

  • 8 N/m2

     

(3)

F=α+βx2

Work done ∫dW=∫F·dx

⇒ △W=∫F·dx=∫(α+βx2)dx

⇒ △W=|αx+βx33|01=α+β3=5

Given α = 1

So, β3=4 ⇒ β=12 N/m2



Q 8 :

A block of mass 25 kg is pulled along a horizontal surface by a force at an angle 45° with the horizontal. The friction coefficient between the block and the surface is 0.25. The block travels at a uniform velocity. The work done by the applied force during a displacement of 5 m of the block is:          [2025]

  • 970 J

     

  • 735 J

     

  • 245 J

     

  • 490 J

     

(3)

N=mg–F2

Block travels with uniform velocity

F2=μ[mg–F2]

F2=0.25[25×9.8–F2]

⇒ 1.25F2=61.25

⇒ F=61.25×21.25=492 N

Wext=FS cos 45°=492×5×12=245 J.



Q 9 :

A force f=x2yi^+y2j^ acts on a particle in a plane x + y = 10. The work done by this force during a displacement from (0, 0) to (4 m , 2 m) is _____ Joule (round off to the nearest integer)          [2025]



(152)

y = 10 – x

W=∫04x2(10–x)dx+∫02y2dy

=[10x33–x44]04+[y33]02=6403–64+83=152.



Q 10 :

Identify the correct statements from the following:                                              [2023]

(A) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket is negative.

(B) Work done by gravitational force in lifting a bucket out of a well by a rope tied to the bucket is negative.

(C) Work done by friction on a body sliding down an inclined plane is positive.

(D) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity is zero.

(E) Work done by the air resistance on an oscillating pendulum is negative.

Choose the correct answer from the options given below:

  • B and E only

     

  • A and C only

     

  • B, D and E only

     

  • B and D only

     

(1)

B. Work done by gravitation will be negative if something is lifted upward.

E. Work done by air resistance is negative.



Q 11 :

A small particle moves to position 5i^-2j^+k^ from its initial position 2i^+3j^-4k^ under the action of force 5i^+2j^+7k^N. The value of work done will be _________ J.           [2023]



(40)

W=F→ (r→f-r→i)

     =(5i^+2j^+7k^)·[(5i^-2j^+k^)-(2i^+3j^-4k^)]

W=40 J



Q 12 :

A force F=(5+3y2) acts on a particle in the y-direction, where F is newton and y is in meter. The work done by the force during a displacement from y = 2 m to y = 5 m is _______ J.          [2023]



(132)

F=5+3y2

W=∫25(5+3y2)dy=[5y+3y33]25=132 J



Q 13 :

A force F→=(2+3x)i^ acts on a particle in the x-direction, where F is in newton and x is in meter. The work done by this force during a displacement from x = 0 to x = 4 m, is __________ J.               [2023]



(32)

W=∫24(2+3x)dx =[2x+3x22]04=8+3×8=32 J



Q 14 :

A block of mass 10 kg is moving along x-axis under the action of force F=5x N. The work done by the force in moving the block from x = 2 m to x = 4 m will be ________ J.                        [2023]



(30)

Work done=∫Fdx

∫245x dx=5[x22]24=52 [16-4]=30 J



Q 15 :

A body of mass 2 kg is moving along x-direction such that its displacement as function of time is given by x(t)=αt2+βt+γm, where α=1 m/s2, β=1 m/s, γ=1 m . The work done on the body during the time interval t=2 s to t=3 s is ______ J.   [2026]

  • 42

     

  • 49

     

  • 12

     

  • 24

     

(4)

x(t)=t2+t+1

v(t)=2t+1

a(t)=2

F=4 N

Displacement=x(3)-x(2)

=13-7=6 m

W=F·S=4×6=24 J



Q 16 :

Two springs have their force constant as k1 and k2 (k1>k2). When they are stretched by the same force up to equilibrium:

  • No work is done by this force in case of both the springs

     

  • Equal work is done by this force in case of both the springs

     

  • More work is done by this force in case of second spring

     

  • More work is done by this force in case of first spring

     

(3)

F1=F2

∵   w1=12k1x12,    k1x1=k2x2=F

w1=12k1[Fk1]2=12F2k1

∴  w2=12F2k2

∵  k1>k2

∴  w1<w2



Q 17 :

A particle undergoes from position O(0,0,0) to A(a,2a,0) via path y=2x2a in x-y plane under the action of a force which varies with particle's (x,y,z) coordinate as F→=x2yi^+yz2e2zj^-(zx+2y)k^. Work done by the force F→ is: (all symbols have their usual meaning and they are in SI unit.)

  • 4a45

     

  • a45

     

  • a34

     

  • 2a45

     

(4)

F→=x2yi^+yz2e2zj^-(zx+2y)k^

dr→=dxi^+dyj^+dzk^

dw=F→·dr→=x2y dx+yz2e2z dy-(zx+2y)dz

for the given path z=0,  y=2x2a

dw=x2y dx=2x4a dx

W=∫dw=2a∫0ax4 dx=2a45



Q 18 :

Statement – I: Work done by friction on a body sliding down an inclined plane is always positive.

Statement – II: Work done is greater than zero, if angle between force and displacement is acute.

  • Both statement I and statement II are correct

     

  • Statement I is incorrect and statement II is correct

     

  • Statement I is correct and statement II is incorrect

     

  • Both statement I and statement II are incorrect

     

(2)

Work done by friction on inclined plane will be negative

W=Fscosθ=+ve, if θ<90°.



Q 19 :

A mass of 1 kg kept in an inclined plane with 30° inclination with respect to horizontal plane and it is at rest initially. Then the whole assembly is moved up with constant velocity of 4 m/s. The work done by the frictional force in time 2 s is ______ J. (Take g = 10 m/s2)                  [2026]

  • 20

     

  • 25

     

  • 30

     

  • 10

     

(1)

fx=mgsinθ=5 N

[IMAGE 15]

W=fs(vt)sinθ=5×4×2×12=20 N



Q 20 :

A body of mass 1 kg moves along a straight line with a velocity v=2x2. The work done by the body during displacement from x=0 to 5 m is ______ J.        [2026]

  • 0

     

  • 250

     

  • 1250

     

  • 1000

     

(3)

W=ΔK.E=Kf-Ki

W=12m(vf2-vi2)

=12×1[(2(5)2)2-(2(0))2]05

=12×50×50=1250 J



Q 21 :

The velocity at which 6 kg mass (shown in figure) strikes the ground when it is released from a height of 6 m above the ground is ______ m/s. Assume pulley is massless and string is light and inextensible. (Take g = 10 m/s2).                       [2026]

[IMAGE 16]

  • 7.74

     

  • 7.20

     

  • 6.55

     

  • 4.50

     

(1)

[IMAGE 17]

WG=ΔK

mBgh-mAgh=12(mA+mB)v2

(6-2)×10×6=12(8)v2

⇒v=60 m/s=7.746 m/s



Q 22 :

A 1 kg block subjected to two simultaneous forces (2i^+3j^+4k^) N and (3i^-j^-2k^) N is moved a distance of 25 m along (3i^-4j^) direction. The work done in this process is _______ J.                       [2026]



(35)

F→1=(2i^+3j^+4k^) NF→2=(3i^-j^-2k^) N}⇒F→net=(5i^+2j^+2k^) N

(distance) d=25 m along (3i^-4j^) direction

∴  displacement d→=25(3i^-4j^9+16)=25(3i^-4j^5) m

d→=(15i^-20j^) m

So, work done W=F→net·d→

W=(5i^+2j^+2k^)·(15i^-20j^)=35 J