Question

A body of mass 10 kg is kept on a rough horizontal surfa of coefficient of friction 0.3. If a horizontal for of 50 N is applied on the body, then the acleration of the body is (Acleration due to gravity $= 10 \text{ ms}^{-2}$)

• A. $5 \text{ ms}^{-2}$
• B. $2 \text{ ms}^{-2}$
• C. $3 \text{ ms}^{-2}$
• D. $1 \text{ ms}^{-2}$

Hint:

Tip: Compare applied force with maximum static friction. If motion occurs, use kinetic friction for net force calculation.

Answer 1

The Correct Option is B

First, calculate the normal reaction: $N = mg = 10 \cdot 10 = 100 \text{ N}$. The maximum static friction is $\mu N = 0.3 \cdot 100 = 30 \text{ N}$. Since the applied force (50 N) exceeds this, motion occurs, and kinetic friction applies, taken as same ($\mu_k = 0.3$). Net force acting: \[ F_{\text{net}} = 50 - 30 = 20 \text{ N}, \quad a = \frac{F_{\text{net}}}{m} = \frac{20}{10} = 2 \text{ ms}^{-2} \] Hence, the acceleration is $2 \text{ ms}^{-2}$.

Answer 2

Step 1: Identify given data
Mass of body, \(m = 10 \text{ kg}\)
Coefficient of friction, \(\mu = 0.3\)
Applied force, \(F = 50 \text{ N}\)
Acceleration due to gravity, \(g = 10 \text{ ms}^{-2}\)

Step 2: Calculate the normal force \(N\)
Since the body is on a horizontal surface,
\[ N = mg = 10 \times 10 = 100 \text{ N} \]

Step 3: Calculate the frictional force \(F_f\)
\[ F_f = \mu N = 0.3 \times 100 = 30 \text{ N} \]

Step 4: Calculate the net force \(F_{\text{net}}\)
\[ F_{\text{net}} = F - F_f = 50 - 30 = 20 \text{ N} \]

Step 5: Calculate acceleration \(a\)
Using Newton's second law,
\[ a = \frac{F_{\text{net}}}{m} = \frac{20}{10} = 2 \text{ ms}^{-2} \]

Step 6: Conclusion
The acceleration of the body is \(2 \text{ ms}^{-2}\).