Newtons Laws Of Motion

According to question, we can draw the following diagram

m%20g%20%5Csin%20%5Ctheta%20%3D%5Cmu(F%2Bm%20g%20%5Ccos%20%5Ctheta)

F%20%3D%5Cfrac%7Bm%20g%20%5Csin%20%5Ctheta%7D%7B%5Cmu%7D-m%20g%20%5Ccos%20%5Ctheta

%3Dm%20g%5Cleft%5B%5Cfrac%7B%5Csin%20%5Ctheta%7D%7B%5Cmu%7D-%5Ccos%20%5Ctheta%5Cright%5D

Here,

m%20%3D1%20%5C%2Ckg%20%2C%20g%3D10%20m%20%2F%20s%20%5E%7B2%7D%2C%20%5Ctheta%3D30%5E%7B%5Ccirc%7D%2C%20%5Cmu%3D0.2

F%20%5Cleft.%3D1%20%5Ctimes%2010%20%5Cfrac%7B%5B%5Csin%2030%7D%7B02%7D-%5Ccos%2030%5E%7B%5Ccirc%7D%5Cright%5D

%3D10%5Cleft%5B%5Cfrac%7B1%7D%7B2%20%5Ctimes%200.2%7D-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5Cright%5D

%3D10%5Cleft%5B%5Cfrac%7B5%7D%7B2%7D-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5Cright%5D%3D5%5B5-%5Csqrt%7B3%7D%5D

%3D5%5B5-1.732%5D%3D16.34%20N

Step 1: Given Data
- Initial acceleration: $a_1%20%3D%206%20%5C%20ms%5E%7B-2%7D$ - Final acceleration: $a_2%20%3D%2011%20%5C%20ms%5E%7B-2%7D$ - Initial force: $F_1%20%3D%2020%20%5C%20N$ - Final force: $F_2%20%3D%2030%20%5C%20N$ - Acceleration due to gravity: $g%20%3D%2010%20%5C%20ms%5E%7B-2%7D$ - Coefficient of kinetic friction: $%5Cmu_k$ (to be determined)

Step 2: Applying Newton's Second Law
Using the equation of motion: $F_1%20-%20f_k%20%3D%20m%20a_1$ $F_2%20-%20f_k%20%3D%20m%20a_2$ where $f_k$ is the kinetic friction force and is given by: $f_k%20%3D%20%5Cmu_k%20m%20g.$

Step 3: Solving for $%5Cmu_k$
Rearranging the first equation: $20%20-%20%5Cmu_k%20m%20g%20%3D%20m%20(6).$ $20%20-%20%5Cmu_k%20m%20(10)%20%3D%206m.$ $20%20%3D%206m%20%2B%2010%5Cmu_k%20m.$ Dividing throughout by $m$ : $20%2Fm%20%3D%206%20%2B%2010%5Cmu_k.$ Similarly, for the second equation: $30%20-%20%5Cmu_k%20m%20g%20%3D%20m%20(11).$ $30%20-%20%5Cmu_k%20m%20(10)%20%3D%2011m.$ $30%20%3D%2011m%20%2B%2010%5Cmu_k%20m.$ Dividing by $m$ : $30%2Fm%20%3D%2011%20%2B%2010%5Cmu_k.$

Step 4: Finding $%5Cmu_k$
Subtracting both equations: $%5Cfrac%7B30%7D%7Bm%7D%20-%20%5Cfrac%7B20%7D%7Bm%7D%20%3D%20(11%20%2B%2010%5Cmu_k)%20-%20(6%20%2B%2010%5Cmu_k).$ $%5Cfrac%7B10%7D%7Bm%7D%20%3D%205.$ $m%20%3D%202.$ Substituting in $20%2Fm%20%3D%206%20%2B%2010%5Cmu_k$ : $%5Cfrac%7B20%7D%7B2%7D%20%3D%206%20%2B%2010%5Cmu_k.$ $10%20%3D%206%20%2B%2010%5Cmu_k.$ $10%5Cmu_k%20%3D%204.$ $%5Cmu_k%20%3D%200.4.$

Step 5: Conclusion
Thus, the coefficient of kinetic friction is: $%5Cmathbf%7B0.4%7D.$

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Newtons Laws Of Motion

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs