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Kinetic Friction

Learn kinetic friction definition and formula with examples, calculate acceleration with force applied at an angle.

Calculation of Acceleration of Body with Kinetic Friction 

  • Consider a block of mass \(m\) which is moving with initial velocity \(v\) on a rough surface, as shown in figure.

  • As the body is in motion, the kinetic friction will act on it.
  • The coefficient of kinetic friction is \(\mu_k\).

Direction of kinetic frictional force 

It always acts in opposite direction to the relative motion between the two surfaces in contact.

Net Horizontal Force 

       \(f= \mu_k\,N\)

  • Using Newton's Second Law 

     \(a= \dfrac{\sum\,F_{net}}{m}\)

 \(a= \dfrac{\mu_k\,N}{m}\)

 \(a = \dfrac{\mu_k\,mg}{m}\)        (\(\because N=mg\))

\(a= \mu_k\,g\)

 

Illustration Questions

A block of mass \(m= 2\,kg\) is sliding over a rough surface having coefficient of friction \(\mu_k = 0.2\). Find the acceleration of the block. (Given : \(g= 10\,m/s^2\))

A \(2\,m/s^2\)

B \(3\,m/s^2\)

C \(4\,m/s^2\)

D \(5\,m/s^2\)

×

Direction of kinetic frictional force 

It always acts in opposite direction to the relative motion between the two surfaces in contact.

FBD of the block 

Net Horizontal Force 

       \(f= \mu_k\,N\)

Using Newton's Second Law

     \(a= \dfrac{\sum\,F_{net}}{m}\)

 \(a= \dfrac{\mu_k\,N}{m}\)

\(\because N=mg,\)  \(\therefore \,a = \dfrac{\mu_k\,mg}{m}\)

\(a= \mu_k\,g\)  

image

Given : \(g= 10 \,m/s^2,\) \(\mu_k =0.2\)

\(a= 0.2× 10\)

\(a=2\,m/s^2\)

 

A block of mass \(m= 2\,kg\) is sliding over a rough surface having coefficient of friction \(\mu_k = 0.2\). Find the acceleration of the block. (Given : \(g= 10\,m/s^2\))

image
A

\(2\,m/s^2\)

.

B

\(3\,m/s^2\)

C

\(4\,m/s^2\)

D

\(5\,m/s^2\)

Option A is Correct

Calculate of Acceleration with Force Applied at an Angle 

  • Consider a block of mass \(m\) which is pulled by a force \(F\) at an angle \(\theta\) with the horizontal.

FBD of the Block 

  • Here, normal force will not be equal to mg because vertical component of force is also acting.
  • Net Force in Vertical Direction 

\(N + F\,sin \,\theta = mg\)

\(N = mg - F\,sin \,\theta\)     ....(1)

  • Applying Newton's Second law in horizontal direction 

\(F\,cos\,\theta - f_k = m\,a\)

\(F\,cos\,\theta - \mu_k \,N = m\,a\)

\(F\,cos\,\theta - \mu_k (mg-F\,sin\,\theta) =ma\)          [from (1)]

\(a= \dfrac{F\,cos\,\theta-\mu_k(mg-F\,sin\,\theta)}{m}\)

 

Illustration Questions

A block of mass \(m= 1\,kg\) is pulled with a force \(F= 10\, N\) at an angle \(\theta= 37° \) with the horizontal. Find the acceleration of the block if the coefficient of kinetic friction is \(\mu_k = 0.2\). [Given : \(g= 10\,m/s^2\), \(sin\,37° = \dfrac{3}{5}, \,cos\,37° =\dfrac{4}{5}\)]  

A \(5.8\,m/s^2\)

B \(6\,m/s^2\)

C \(7.2\ m/s^2\)

D \(8\,m/s^2\)

×

FBD of the Block 

Here, normal force will not be equal to mg because vertical component of force is also acting.

image

Net Force in Vertical Direction 

   \(N + F\,sin \,\theta = mg\)

\(N = mg - F\,sin \,\theta\)     ....(1)

Applying Newton's Second law in horizontal direction 

\(F\,cos\,\theta - f_k = m\,a\)

\(F\,cos\,\theta - \mu_k \,N = m\,a\)

\(F\,cos\,\theta - \mu_k (mg-F\,sin\,\theta) =ma\)          [from (1)]

\(a= \dfrac{F\,cos\,\theta-\mu_k(mg-F\,sin\,\theta)}{m}\)

Given : \(m= 1\,kg, \,F=10 \,N, \,\theta= 37°, \,\mu_k = 0.2,\,g=10\,m/s^2\)

\(a= \dfrac{(10)cos\,37° - (0.2)[1× 10-(10)sin\,37° ]}{1}\)

\(a = 10\left(\dfrac{4}{5}\right) - (0.2) \left[10-10\left(\dfrac{3}{5}\right)\right]\)

\(a= 8-(0.2)[10-6]\)

\( a= 8-(0.2)[4]\)

\(a= 8-0.8\)

\(a= 7.2\,m/s^2\)

A block of mass \(m= 1\,kg\) is pulled with a force \(F= 10\, N\) at an angle \(\theta= 37° \) with the horizontal. Find the acceleration of the block if the coefficient of kinetic friction is \(\mu_k = 0.2\). [Given : \(g= 10\,m/s^2\), \(sin\,37° = \dfrac{3}{5}, \,cos\,37° =\dfrac{4}{5}\)]  

image
A

\(5.8\,m/s^2\)

.

B

\(6\,m/s^2\)

C

\(7.2\ m/s^2\)

D

\(8\,m/s^2\)

Option C is Correct

Calculation of Distance Traveled with Kinetic Friction 

  • Consider a block of mass \(m\) which is thrown with initial velocity \(v\) on a rough horizontal surface, as shown in figure.
  • The coefficient of kinetic friction between rough surface and block is \(\mu_k\)

  • The  sliding of block on surface will start only when the block comes in contact with surface and kinetic friction comes into act to oppose it.
  • As the block is sliding towards right, thus to oppose it, the kinetic frictional force will act towards left.
  • Applying Newton's second law in vertical direction 

         \(N =mg\)  ........(1)

  • Applying Newton's second law in horizontal direction 

         \(- f_k = m\,a\)         [taking right side positive]

\(a = \dfrac{-f_k}{m}\)

\(a= \dfrac{-\mu_k\,mg}{m}\)

\(a= -\mu_k\,g\)

  • Finally, the block will come to rest.

           Thus,   \(v=0\) 

using second law of motion,

\(v^2 =u^2 +2\,a\,s\)

\(0 = u^2 -2\,\mu_k \,g\,s\)

\(s= \dfrac{u^2}{2\,\mu_k\,g}\)

where, \(s\) is the distance traveled by block before it comes to rest.

Illustration Questions

A block of mass \(m=1\,kg\) is thrown with velocity \(v= 10\,m/s\) on a rough horizontal surface, having coefficient of kinetic friction \(\mu_k = 0.1\), as shown in figure. What will be the distance traveled by the block, before it comes to rest? [Given : \(g= 10 \,m/s^2\)]

A 20 m

B 25 m

C 50 m

D 30 m

×

The  sliding of block on surface will start only when the block comes in contact with surface and kinetic friction comes into act to oppose it.

As the block is sliding towards right, thus to oppose it the kinetic frictional force will act towards left.

FBD of the Block 

Applying Newton's second law in vertical direction 

         \(N =mg\)  ........(1)

Applying Newton's second law in horizontal direction 

         \(- f_k = m\,a\)

[taking right side positive]

\(a = \dfrac{-f_k}{m}\)

\(a= \dfrac{-\mu_k\,mg}{m}\)

\(a= -\mu_k\,g\)  

image

Finally, the block will come to rest

           Thus, \(v=0\) 

using second law of motion, 

\(v^2 =u^2 +2\,a\,s\)

\(0 = u^2 -2\,\mu_k \,g\,s\)

\(s= \dfrac{u^2}{2\,\mu_k\,g}\)

where, \(s\) is the distance traveled by block before it comes to rest

Given : \(u= 10\,m/s , \, \mu_k = 0.1 , \,g= 10\,m/s^2\)

\(s=\dfrac{(10)^2}{2× (0.1)(10)}\)

\(s=50\,m\)

 

A block of mass \(m=1\,kg\) is thrown with velocity \(v= 10\,m/s\) on a rough horizontal surface, having coefficient of kinetic friction \(\mu_k = 0.1\), as shown in figure. What will be the distance traveled by the block, before it comes to rest? [Given : \(g= 10 \,m/s^2\)]

image
A

20 m

.

B

25 m

C

50 m

D

30 m

Option C is Correct

Calculation of Maximum Acceleration of a Flatbed Truck so that Block Placed on it does not Slip

  • Consider a flatbed truck which is standing on a horizontal road.
  • A block of mass \(m\) is kept on the truck, as shown in figure.

  • Now, the truck accelerates. The block will remain, as it is till maximum limit of acceleration is achieved.
  • If the truck accelerates above this limits, the block may slip.
  • Consider that the truck moves towards right side.
  • At the verge of slipping, the acceleration of both truck and block is same and here, limiting friction must be acting.
  • Limiting Friction 

         It is the maximum value of static friction i.e., \(f_{max} = \mu_s \,N\) 

where \(\mu_s\) is the  coefficient of static friction. 

  • As truck accelerates towards right direction, the block will have tendency to slip towards left. So, frictional force on the block will act towards right direction.
  • Applying Newton's Second law in vertical direction 

          \(N = mg\) 

{Equilibrium in vertical direction }

  • Applying Newton's Second law in horizontal direction 

          \(f_{max} = m\,a_{max}\)

[At the verge of slipping \(f_{max}\) is acting]

          \(\Rightarrow \mu_s\,N = m\,a_{max}\) 

         \(\Rightarrow \mu_s\,mg=ma_{max}\)   

         \(\Rightarrow a_{max} = \mu_s\,g\)

This is the maximum value of acceleration of truck so that block does not slip.

Illustration Questions

A flatbed truck is moving at a speed of \(v= 10\,m/s\). A block is kept on it, as shown in figure. The coefficient of friction between block and surface of truck is \(\mu= 0.5\). Calculate the maximum limits of acceleration of truck, so that block does not slip. [Given : \(g=10\,m/s^2\)] 

A \(4\,m/s^2\)

B \(5\,m/s^2\)

C \(3\,m/s^2\)

D \(2\,m/s^2\)

×

Equilibrium in upper direction

\(N= mg\)

 

 

 

image

Just before slipping, the value of limiting friction

\(f_L = f_{max} = \mu\,N\)

 

At the verge of slipping, applying Newton's Second law,

\(m\,a = f_L\)

\(\Rightarrow m\,a = \mu\,N\)

\(\Rightarrow m\,a = \mu\,mg\)

\(\Rightarrow \,a _{max}= \mu\,g\)

 

Given : \(\mu = 0.5, \,\,g= 10\,m/s^2\)

\(a_{max}= 0.5× 10 \)

\(\Rightarrow a_{max}= 5\,m/s^2\)

 

A flatbed truck is moving at a speed of \(v= 10\,m/s\). A block is kept on it, as shown in figure. The coefficient of friction between block and surface of truck is \(\mu= 0.5\). Calculate the maximum limits of acceleration of truck, so that block does not slip. [Given : \(g=10\,m/s^2\)] 

A

\(4\,m/s^2\)

.

B

\(5\,m/s^2\)

C

\(3\,m/s^2\)

D

\(2\,m/s^2\)

Option B is Correct

Acceleration of Blocks Placed on Wedge 

  • Consider two blocks A and B of masses m each, are attached with a string.
  • Block A lies on the rough surface of a fixed wedge with an angle of inclination \(\theta\), while block B hangs through a massless pulley, as shown in figure. 

  • The coefficient of static and kinetic friction between the surface of wedge and block A are \(\mu_s\) and \(\mu_k\), respectively.
  • Representation of forces on the system 

FBD of Block A   

\(N = mg \,cos\,\theta\)

 

 

FBD of Block B   

      \(T = mg\)    .......(1)

  • Maximum frictional force 

    \(f_s = f_{max} = \mu_s\,N= \mu_s \,mg\,cos\,\theta\)

  • Kinetic friction 

      \(f_k = \mu_kN = \mu_k \,mg \,cos\,\theta\)    ......(2) 

  • As the mass of block B is equal to the mass of block A, thus, assume that there is no slipping of block A.

                \(a_A =0\)

  • Direction of frictional force 

       To find the direction of frictional force, first find the tendency of relative motion.

         As  \(T > mg\,sin\,\theta\)

\([\because \,T =mg \,\,\,{\text{and}} \,\,\theta\neq 90°]\)

Thus, block A has tendency to move upwards.

Hence, the frictional force \(f\) will act downwards.

  • Since, acceleration of block A is zero thus, it will not move.

Applying Newton's Second Law on block A 

      \(mg\,sin\,\theta +f -T=0\)

        \(f= T- mg\,sin\,\theta\) 

  • Case 1 :   If \(f< f_{max}\)  

    Then, assumed direction of frictional force is correct.

  • Case 2 :    If  \(f>f_{max}\) 

Then, it is not possible.

Hence, there must be some acceleration and the block A will move in upward direction.

  • The block A will, move with some acceleration.

            Thus, kinetic frictional force will act.

  • Using Newton's second law 

            \(T- f_k - mg\,sin\,\theta = m\,a\)

            \(a= \dfrac{T- f_k -mg\,sin\,\theta}{m}\)

    \(a= \dfrac{mg- \mu_k \,mg \,cos\,\theta - mg \,sin\,\theta }{m}\)

[ From (1) and (2) ]

 \(a= g[1 -\mu_k \,cos\,\theta -sin\,\theta ]\)

Illustration Questions

Two blocks A and B of mass \(m= 10 \,kg\) each are attached with a string. The block A lies on a wedge having angle of inclination \(\theta = 37° \), while block B hangs through a massless pulley, as shown in figure. Find the acceleration of the blocks, if coefficient of static and kinetic friction between the surface of wedge and block A are \(\mu_s= 0.2\,\,{\text{and}}\,\,\mu_k = 0.1\) respectively. [Given : \(g= 10\,m/s^2\),  \(sin\,37° = \dfrac{3}{5} , \,\,cos\,37° = \dfrac{4}{5}\)]  

A \(4\,m/s^2\)

B \(2.3\,m/s^2\)

C \(3.2\,m/s^2\)

D \(6\,m/s^2\)

×

FBD of Block A 

\(N = mg \,cos\,\theta\)

image image

FBD of Block B

      \(T = mg\)    .......(1)

       \(N = (10)(10) \,cos\,37­° \)

      \(N = 10× 10 × \dfrac{4}{5} = 80 \,N\)

       \(T = 100\,N\)   

image image

Maximum frictional force

    \(f_s = f_{max} = \mu_s\,N= \mu_s \,mg\,cos\,\theta\)

Kinetic friction 

      \(f_k = \mu_kN = \mu_k \,mg \,cos\,\theta\)    ......(2)  

    \(f_{max} = (0.2) 80 =16 \,N\)

    \(f_{k} = (0.1) 80 =8 \,N\)

 

image

As the mass of block B is equal to the mass of block A, Thus, assume that there is no slipping of block A.

\(a_A =0\)

image

Direction of frictional force

To find the direction of frictional force, first find the tendency of relative motion.

As  \(T > mg\,sin\,\theta\)

\([\because \,T =100\,N\,\,\,{\text{and}} \,\,mg\,sin\,\theta=60\,N]\)

Thus, block A has tendency to move upwards

Hence, the frictional force \(f\)will act downwards    

image

Since, acceleration of block A is zero thus, it will not move.

Applying Newton's second law on block A 

\(mg\,sin\,\theta +f -T=0\)

\(f= T- mg\,sin\,\theta\)  

         \(f= 100-60 =40 \,N\)

 

image

 If  \(f>f_{max}\) 

Then, it is not possible.

Hence, there must be some acceleration and the block A will move in upward direction.

image

The block A will move with some acceleration

Thus, kinetic frictional force will act.

Using Newton's second law,

 \(T- f_k - mg\,sin\,\theta = m\,a\)

 \(a= \dfrac{T- f_k -mg\,sin\,\theta}{m}\)

 \(a= \dfrac{mg- \mu_k \,mg \,cos\,\theta - mg \,sin\,\theta }{m}\)

 from (1) and (2)

\(a= g[1 -\mu_k \,cos\,\theta -sin\,\theta ]\)  

image

Given : \(g= 10\,m/s^2 , \,\mu_k = 0.1,\,\theta = 37° \)

\(a= 10[1-(0.1)\,cos\,37° -sin\,37°]\)

\(a= 10 \left[1-(0.1)\dfrac{4}{5}- \dfrac{3}{5}\right]\)

\(a= 10 \left[\dfrac{5-0.4-3}{5}\right]\)

\(a= 2 \left[1.6\right]\)

\(a= 3.2\,m/s^2\)

 

image

Two blocks A and B of mass \(m= 10 \,kg\) each are attached with a string. The block A lies on a wedge having angle of inclination \(\theta = 37° \), while block B hangs through a massless pulley, as shown in figure. Find the acceleration of the blocks, if coefficient of static and kinetic friction between the surface of wedge and block A are \(\mu_s= 0.2\,\,{\text{and}}\,\,\mu_k = 0.1\) respectively. [Given : \(g= 10\,m/s^2\),  \(sin\,37° = \dfrac{3}{5} , \,\,cos\,37° = \dfrac{4}{5}\)]  

image
A

\(4\,m/s^2\)

.

B

\(2.3\,m/s^2\)

C

\(3.2\,m/s^2\)

D

\(6\,m/s^2\)

Option C is Correct

Calculation of Maximum Value of Applied Force

  • Consider two blocks of masses m1 and m2, placed on each other, as shown in figure.

  • The coefficient of friction between the blocks is \(\mu\).
  • An external force \(F_{ext}\) is applied on the block of mass m1.
  • To calculate the maximum value of F such that blocks of mass m1 and m2 could move together, firstly we have to calculate the frictional force acting between the blocks.
  • At the verge of slipping, the limiting frictional force \(f_{\ell}\) acts between the blocks.

FBD of block of mass m1

Limiting frictional force,  \(f_{\ell} = \mu\,N\)

\(f_{\ell } = \mu\,m_1\,g\)    ........(1) 

\([\because N= m_1 \,g]\)

FBD of block of mass m2

Let the acceleration of both the blocks are \(a_{\ell}\).

Applying Newton's Second Law on block m2 

   \(f_{\ell} = m_2 \,a_{\ell}\)

    \( a_{\ell} = \dfrac{f_{\ell}}{m_2} \,\)

    \(a_{\ell} = \dfrac{\mu \,m_1\,g}{m_2} \,\)

 

[from equation (1)]

     \(a_{\ell} = \mu \left(\dfrac{\,m_1}{m_2}\right)\,g\)       .......(2)  

  • Applying Newton's Second law on block m1 

\(F_{max} - f_{\ell} = m_1\,a_{\ell}\)

\(F_{max} = m_1\left(\dfrac{\mu\,m_1\,g}{m_2}\right) +\mu\,m_1 \,g\)

[from (1)and (2)]

 \(F_{max} =\dfrac{\mu\,{m_1}^2\,g}{m_2} +\mu\,m_1 \,g\)

\(F_{max} = \mu\,m_1 \,g \left[\dfrac{m_1+m_2}{m_2}\right] \)

 

  • Hence, \(F_{max}\) is the value of maximum external force for which both the blocks move together.
  • It is also clear that  \(F_{max}\neq f_{\ell}\).

Illustration Questions

Two blocks of masses \( m_1 = 5\, kg\) and \( m_2 = 10\, kg\) are placed on each other, as shown in figure. The coefficient of friction between them is \(\mu=0.5\). Calculate the maximum value of force which is applied on the block of mass \(m_1\) such that both blocks could move together. [Given :- \(g= 10\,m/s^2\)]

A \(40 \,N\)

B \(37.5\,N\)

C \(42 \,N\)

D \(44 \,N\)

×

FBD of block of mass m1

Limiting frictional force, \(f_{\ell} = \mu\,N\)

\(f_{\ell } = \mu\,m_1\,g\)    ........(1)

\([\because N= m_1 \,g]\)

image

FBD of block of mass m2

Let the acceleration of both the blocks is \(a_{\ell}\).

Applying Newton's Second Law on block m2  

    \(f_{\ell} = m_2 \,a_{\ell}\)

       \(a_{\ell} = \dfrac{f_{\ell}}{m_2} \,\)

       \(a_{\ell} = \dfrac{\mu \,m_1\,g}{m_2} \,\)

 

image

[from equation (1)]

 \(a_{\ell} = \mu \left(\dfrac{\,m_1}{m_2}\right)\,g\)       .......(2)  

image

Applying Newton's Second law on block m1 

\(F_{max} - f_{\ell} = m_1\,a_{\ell}\)

\(F_{max} = m_1\left(\dfrac{\mu\,m_1\,g}{m_2}\right) +\mu\,m_1 \,g\)

[from (1) and (2)]

 \(F_{max} =\dfrac{\mu\,{m_1}^2\,g}{m_2} +\mu\,m_1 \,g\)

\(F_{max} = \mu\,m_1 \,g \left[\dfrac{m_1+m_2}{m_2}\right] \)   

image

Given : \(\mu = 0.5 , \, m_1 = 5\,kg,\,m_2 = 10 \,kg , \,g = 10 \,m/s^2\)

\(F_{max} = (0.5) \,5 × 10 \left[\dfrac{5+10}{10}\right] \)

\(F_{max} = 25 \left[\dfrac{15}{10}\right]\)

\(F_{max} = \left[\dfrac{75}{2}\right]\)

\(F_{max} = 37.5\,N\)

Two blocks of masses \( m_1 = 5\, kg\) and \( m_2 = 10\, kg\) are placed on each other, as shown in figure. The coefficient of friction between them is \(\mu=0.5\). Calculate the maximum value of force which is applied on the block of mass \(m_1\) such that both blocks could move together. [Given :- \(g= 10\,m/s^2\)]

image
A

\(40 \,N\)

.

B

\(37.5\,N\)

C

\(42 \,N\)

D

\(44 \,N\)

Option B is Correct

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