Question

A block is kept on a rough horizontal surface. The acceleration of the block increases from \( 6 \ ms^{-2} \) to \( 11 \ ms^{-2} \) when the horizontal force acting on it increases from \( 20 \ N \) to \( 30 \ N \). The coefficient of kinetic friction between the block and the surface is: (Acceleration due to gravitiy = 10ms-^2)

• A. \( 0.2 \)
• B. \( 0.3 \)
• C. \( 0.4 \)
• D. \( 0.5 \)

Hint:

To solve friction problems involving Newton's second law, set up force balance equations for different conditions and solve for the unknown friction coefficient.

Answer 1

The Correct Option is C

Step 1: Given Data 
- Initial acceleration: \( a_1 = 6 \ ms^{-2} \) - Final acceleration: \( a_2 = 11 \ ms^{-2} \) - Initial force: \( F_1 = 20 \ N \) - Final force: \( F_2 = 30 \ N \) - Acceleration due to gravity: \( g = 10 \ ms^{-2} \) - Coefficient of kinetic friction: \( \mu_k \) (to be determined)

 Step 2: Applying Newton's Second Law 
Using the equation of motion: \[ F_1 - f_k = m a_1 \] \[ F_2 - f_k = m a_2 \] where \( f_k \) is the kinetic friction force and is given by: \[ f_k = \mu_k m g. \] 

Step 3: Solving for \( \mu_k \) 
Rearranging the first equation: \[ 20 - \mu_k m g = m (6). \] \[ 20 - \mu_k m (10) = 6m. \] \[ 20 = 6m + 10\mu_k m. \] Dividing throughout by \( m \): \[ 20/m = 6 + 10\mu_k. \] Similarly, for the second equation: \[ 30 - \mu_k m g = m (11). \] \[ 30 - \mu_k m (10) = 11m. \] \[ 30 = 11m + 10\mu_k m. \] Dividing by \( m \): \[ 30/m = 11 + 10\mu_k. \] 

Step 4: Finding \( \mu_k \) 
Subtracting both equations: \[ \frac{30}{m} - \frac{20}{m} = (11 + 10\mu_k) - (6 + 10\mu_k). \] \[ \frac{10}{m} = 5. \] \[ m = 2. \] Substituting in \( 20/m = 6 + 10\mu_k \): \[ \frac{20}{2} = 6 + 10\mu_k. \] \[ 10 = 6 + 10\mu_k. \] \[ 10\mu_k = 4. \] \[ \mu_k = 0.4. \]

 Step 5: Conclusion 
Thus, the coefficient of kinetic friction is: \[ \mathbf{0.4}. \]