Informative line

Work Energy Theorem

Learn work-kinetic energy theorem problems with examples and change in kinetic energy, practice equation to calculate work done by friction, gravity, constant force, spring force, force of gravity.

Work-Kinetic Energy Theorem 

  • When work is done on a system, its speed changes. 
  • Due to the change in its speed, the kinetic-energy of the system changes.

Thus,  \(W = \Delta K E\)

This is known as work-Kinetic energy theorem.

  • If the net work done on a system is positive, then the speed of the system increases. 

As \((KE)_{final \;} >\;(KE)_{initial }\)

Then, work done,

\(W = \Delta K E\)

\(W = (KE)_{final}-(KE)_{initial }\)

\(W = Positive\)

  •  Similarly, if the work done on a system is negative, then the speed of the system decreases. 

As \((KE)_{final}\; <\; (KE) _{initial }\)

Then, work done, \(W = \Delta KE\)

\(W = (KE)_{final } - (KE) _{initial }\)

\(W = Negative \)

Illustration Questions

Under the application of a certain force, the speed of the system is continuously decreasing. The work done by the force is

A Positive 

B Negative 

C Zero 

D  Increasing 

×

According to work-kinetic- energy theorem, \(W = \Delta KE\)

Since, speed is decreasing,

thus,  \(v_f\; < \; v_i\)

\(\Rightarrow\; (KE)_f \; < \; (KE)_i \)

Work done, \(W = (KE)_f - (KE)_i\)

W = Negative 

Under the application of a certain force, the speed of the system is continuously decreasing. The work done by the force is

A

Positive 

.

B

Negative 

C

Zero 

D

 Increasing 

Option B is Correct

Calculation of Work Done by Force of Gravity

Case-1: When a particle is projected in vertically upward direction 

  • Consider a particle of mass m which is projected in vertically upward direction with some velocity v.
  • When particle reaches at its maximum height, its velocity becomes zero.
  • Thus, v=v and vf = 0

Change in kinetic energy

\(\Delta KE = (KE)_f - (KE)_i\)

\(\Delta KE = \dfrac {1}{2} × m × v_f^2- \dfrac{1}{2}× m ×v_i^2 \)

\(\Delta KE= \dfrac{1}{2}× m × (0)^2 - \dfrac{1}{2}× mv^2\)

\(\Delta KE = \dfrac{-1}{2} mv^2\)

  • By work-energy theorem 

\(W = \Delta KE\)

\(W = \dfrac{-1}{2}mv^2\)

Thus, work done is negative when particle is thrown vertically upward.

  • This can also be verified by definition of work.

  • When particle is projected vertically upward, then both the displacement (s) and force of gravity (Fg) are in opposite direction, as shown in figure.  
  • By definition of work, if force of gravity and displacement are in opposite direction, then work done by force is negative.
  • Hence, it is verified that when particle is projected vertically upward, then work done is negative. 

Case-2: When particle is dropped (free fall) 

  • Consider a particle of mass m which is dropped from some height.
  • In free fall, when particle is dropped vertically downward, then its initial velocity is zero. 
  • During free fall, the particle will gain some velocity v, i.e., final velocity of particle. 
  • Hence, vi = 0 and vf = v

Change in kinetic energy  

\(\Delta KE = (KE)_f- (KE)_i \)

\(\Delta KE = \dfrac{1}{2} m (v_f)^2-0\)

\(\Delta KE = \dfrac{1}{2} mv^2 \)

By work-energy theorem

W = \(\Delta\) KE

W = \(\dfrac{1}{2}mv^2\) 

  • Thus, in free fall of an object, the work done by gravity is positive.
  • Positive work done can be verified by the definition of work.

  • In free fall of an object, both displacement (s) and force of gravity (Fg) are in same direction.
  • By definition of work, if the force of gravity and displacement are in same direction, then the work done by force is positive.
  • Hence, it is verified that when particle is dropped vertically downward, then the work done is positive.

Illustration Questions

A ball of mass m = 2 kg is projected in vertically upward direction with initial velocity v = 3 m/sec. Find the work done by gravity, when particle reaches its maximum height. 

A – 20 J

B – 9 J

C – 30 J

D – 6 J

×

Calculation of initial kinetic energy 

\((KE)_i=\dfrac{1}{2}mv_i^2\)

\(\dfrac{1}{2}\)(2)(3)2

= 9 J

Calculation of final kinetic energy,

At maximum height, the velocity of particle becomes zero.

⇒ vf = 0

\((KE)_f=\dfrac{1}{2}\) \(m\;v_f^2\)

(KE)f = 0

By work-kinetic energy theorem

W = \(\Delta KE\)

\(W = (KE)_f – (KE)_i\)

W = 0 – 9

W = – 9 J

A ball of mass m = 2 kg is projected in vertically upward direction with initial velocity v = 3 m/sec. Find the work done by gravity, when particle reaches its maximum height. 

A

– 20 J

.

B

– 9 J

C

– 30 J

D

– 6 J

Option B is Correct

Work Done by Friction 

  • Consider a box of mass m which is given some initial velocity v on a rough surface.
  • After some displacement of the box, the box will stop.
  • If the coefficient of frictional force is \(\mu_ k\), then the work done by frictional force is \(W = \Delta KE\).

  • Initial velocity of the box is v m/s.
  • Initial kinetic energy of the box is 

           (KE)i = \(\dfrac{1}{2}\) mv2

  • Final speed of the box is zero.
  • Final kinetic energy of the box is 

          (KE)f = \(\dfrac{1}{2}\) m(0)= 0

  • Work done by frictional force

W = \(\Delta\) KE

W = \(\left(\dfrac{1}{2}mv^2\right)_f-\left(\dfrac{1}{2}mv^2\right)_i\)

W = \(0-\dfrac{1}{2}mv_i^2\)

W = \(\dfrac{-1}{2}mv^2\)

  • Thus, the work done by frictional force is negative.
  • By above steps, we are able to find change in kinetic energy and we can find work done without knowing friction coefficient.
  • It can also be verified by definition of work.

  • The box is given some velocity and it moves as, shown in figure.   
  • The box displaced in right direction. 
  • The direction of frictional force is in left direction.
  • By the definition of work, if the frictional force and displacement are in opposite direction, then the work done is negative. 

Illustration Questions

A box of mass m = 10 kg is given initial speed vi = 3m/sec on a rough surface. After covering some distance, the block is stopped due to frictional force. Find the work done by frictional force. 

A – 15 J

B 60 J

C – 45 J

D 50 J

×

Calculation of initial kinetic energy

\((KE)_i=\dfrac{1}{2}\)\(mv_i^2\)

\(\dfrac{1}{2}\) × 10 × (3)2

= 45 J

Calculation of final kinetic energy,

Final velocity,vf = 0

Thus, \((KE)_f=\dfrac{1}{2}mv_f^2\)

\((KE)_f=\dfrac{1}{2}m(0)^2 \)

(KE)f = 0 

image

By work-energy theorem,

W = \(\Delta KE\)

\(W = (KE)_f – (KE)_i\)

W = 0 – 45 J

W = – 45 J 

A box of mass m = 10 kg is given initial speed vi = 3m/sec on a rough surface. After covering some distance, the block is stopped due to frictional force. Find the work done by frictional force. 

image
A

– 15 J

.

B

60 J

C

– 45 J

D

50 J

Option C is Correct

Calculation of Work Done by Spring Force

Direction of spring force 

  • Direction of spring force is always towards its natural length. 
  • Force of spring is given by

 \(F_s= -k\;x\)

Where negative sign shows the direction of spring force.

  • Hence, force of spring is a variable force.

  • Consider a block, attached with a spring.
  • The block is displaced from initial position xi to final position x, as shown in figure.

  • As   \(dw = F.dx\)

w = \(\int\limits^f_iF.dx\)

w = \(\int\limits^{x_f}_{x_i} - kx\;dx\)

w = – k \(\left[\dfrac{x^2}{2}\right]^{x_f}_{x_i}\)

w = \(\dfrac{-1}{2}k\;x_f^2\;+\dfrac{1}{2}k\;x_i^2\)

w = \(\dfrac{1}{2}k\;x_i^2\;-\;\dfrac{1}{2}k\;x_f^2\)

When \(x_f\;>\; x_i\) then, work done is negative.

When \(x_i\;>\; x_f\) then, work done is positive.

Case 1. Work done during extension

Here, the displacement (s) and force of spring (Fs) are in opposite direction. Thus, work done is negative. 

Case 2. Work done during release after extension

Here, the displacement (s) and force of spring (Fs) are in same direction. Thus, work done is positive. 

Case 3. Work done during compression

Here, the displacement (s) and force of spring (Fs) are in opposite direction. Thus, work done is negative. 

Case 4. Work done during release after compression

Hence, the displacement (s) and force of spring (Fs) are in same direction. Thus, work done is positive. 

Illustration Questions

A block 'A' is attached to the free end of a spring whose natural length is \(\ell\) = 50 cm. Now the block is displaced from \(x_i\) = 50 cm to \(x_f\) = 60 cm by stretching the spring. Calculate the work done by spring force in stretching. The value of spring constant is k = 80 N/m.

A – 2.2 J

B 3.3 J

C – 4.4 J

D – 1.1 J

×

Work done by spring force in stretching

\(W =\dfrac{1}{2}Kx_i^2 \;- \;\dfrac{1}{2}Kx_f^2\)

W = \(\dfrac{1}{2} × 80 × \left\{ (50 × 10^{-2})^2 - (60 × 10^{-2})^{2}\right\}\)

W = 40 × \(\left\{ - 1100 × 10^{-4}\right\}\)

W = – 44000 × 10–4

W = – 4.4 J

A block 'A' is attached to the free end of a spring whose natural length is \(\ell\) = 50 cm. Now the block is displaced from \(x_i\) = 50 cm to \(x_f\) = 60 cm by stretching the spring. Calculate the work done by spring force in stretching. The value of spring constant is k = 80 N/m.

image
A

– 2.2 J

.

B

3.3 J

C

– 4.4 J

D

– 1.1 J

Option C is Correct

Relation between the Work Done by Friction and Work Done by Gravity

  • Consider a block of mass m which is made to slide over an inclined plane, as shown in figure. 
  • The angle of inclination of the inclined plane is \(\theta\) with the horizontal. 
  • The block slides on the inclined surface upto \(x\) distance and then stops due to the friction force between the surface. The coefficient of friction of rough surface is \(\mu_k \).

  • Change in kinetic energy of the block 

\((KE)_f-(KE)_i\;=\Delta KE\;[\because v_f= 0, v_i =0]\)

\(0\,-\;0\,=\,\Delta KE\)

\(\Delta KE = 0\)

  • Thus, net work done = work done by friction + work done by gravity  

Wnet  =  Wfriction  +  Wgravity  

  • According to work-kinetic energy theorem 

Wnet = \(\Delta\) KE

Wfriction  +  Wgravity = 0

Wgravity =  –Wfriction

\(mg\,sin\,\theta\times x=-(-\mu_kmg\,cos\,\theta)\)

\(x\,sin\,\theta=\mu _kcos\,\theta\)

Illustration Questions

A block of mass m = 1 kg is made to slide over an inclined plane, as shown in figure. If the block slides upto \(x\) = 2 m on the rough surface of the inclined plane find the coefficient of friction, if the angle of inclination is \(\theta\) = 37°.  Given :-  sin 37° = \(\dfrac{3}{5}\)                cos 37° = \(\dfrac{4}{5}\)

A 0.5

B 1.5

C 0.6

D 2.6

×

Change in kinetic energy of the block

\((KE)_f-(KE)_i=\Delta KE\;[\because v_f= 0, v_i =0]\)

\(0-0\;=\Delta KE\)

 

image

Thus, net work done = work done by friction + work done by gravity  

Wnet  = Wfriction + Wgravity  

image

According to work- kinetic energy theorem,

Wnet = \(\Delta\) KE

Wfriction + Wgravity  =  0

Wgravity = – Wfriction

\(mg\,sin\theta\times x=-(-\mu _k\,mg\,cos\theta )\)

\(x\,sin\theta=\mu_kcos\,\theta\)

image

2 × sin 37° = \(\mu _k\) cos 37° 

2 × \(\dfrac{3}{5}\) = \(\mu _k\) × \(\dfrac{4}{5}\)

\(\dfrac{3}{2}\) = \(\mu _k\)

\(\Rightarrow\mu _k=1.5\)

image

A block of mass m = 1 kg is made to slide over an inclined plane, as shown in figure. If the block slides upto \(x\) = 2 m on the rough surface of the inclined plane find the coefficient of friction, if the angle of inclination is \(\theta\) = 37°.  Given :-  sin 37° = \(\dfrac{3}{5}\)                cos 37° = \(\dfrac{4}{5}\)

image
A

0.5

.

B

1.5

C

0.6

D

2.6

Option B is Correct

Relation between the Work Done by Spring Force and Work Done by Applied Constant Force

  • Consider, a block is attached to the free end of a spring, as shown in figure. 
  • Initially, the block is attached to the unstretched configuration of the spring.

  • A constant force \(\vec{F_c}\) is applied applied on the block. 
  • Due to this force, the block gets displaced from its position.
  • By work energy theorem, Wnet = \(\Delta KE\)

    Work done by constant force + work done by spring force = \(\Delta KE\)

    Wc + Ws = \(\Delta KE\)

  • Initially and finally, the block is at rest, thus \(\Delta KE\)= 0
  • Hence,  

 Wc + W= 0

\(W_s=-W_c\)

Illustration Questions

A block A is attached to the free end of a spring whose natural length is \(\ell\) = 50 cm. A constant force is applied on the block. Due to which the block gets displaced from \(x_i\) = 50 cm to \(x_f\) = 60 cm. Calculate the work done by constant force. The value of spring constant is k = 80 N/m. 

A 4.4 J

B 3.2 J

C 4.8 J

D 9 J

×

Work done by spring force in stretching,

\(W =\dfrac{1}{2}Kx_i^2 \;\;- \;\;\dfrac{1}{2}Kx_f^2\)

 

image

Ws\(\dfrac{1}{2} × 80 × \left\{ (50 × 10^{-2})^2 - (60 × 10^{-2})^{2}\right\}\)

Ws = 40 × \(\left\{ - 1100 × 10^{-4}\right\}\)

Ws = – 44000 × 10–4

Ws = – 4.4 J

image

By work energy theorem, Wnet = \(\Delta KE\)

Work done by constant force + work done by spring force = \(\Delta KE\)

Wc + Ws = \(\Delta KE\)

image

Initially and finally, the block is at rest, thus \(\Delta KE\)= 0

image

Hence, Wc + W= 0

\(W_s=-W_c\)

image

Wc = – ( – 4.4 J ) 

Wc = 4.4 J

image

A block A is attached to the free end of a spring whose natural length is \(\ell\) = 50 cm. A constant force is applied on the block. Due to which the block gets displaced from \(x_i\) = 50 cm to \(x_f\) = 60 cm. Calculate the work done by constant force. The value of spring constant is k = 80 N/m. 

image
A

4.4 J

.

B

3.2 J

C

4.8 J

D

9 J

Option A is Correct

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