A 60kg is pushed horizontaly with just enough force to start it moving across a floor and the same force continues to act afterwards. The coefficient of static friction and sliding friction are 0.5and 0.4 respectively the accleration of the body is

Options

A. `6m//s^(2)`
B. `2m//s^(2)`
C. `3m//s^(2)`
D. `1m//s^(2)`

Answer

To solve the problem, we need to determine the acceleration of a 60 kg object being pushed horizontally across a floor, given the coefficients of static and kinetic friction.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the object (m) = 60 kg
- Coefficient of static friction (μ_s) = 0.5
- Coefficient of kinetic friction (μ_k) = 0.4
- Acceleration due to gravity (g) = 9.8 m/s² (approximately 10 m/s² for simplicity)

2. **Calculate the Normal Force (N):**
- The normal force for an object resting on a horizontal surface is equal to its weight.
\[ N = m \cdot g = 60 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 600 \, \text{N} \]

3. **Calculate the Maximum Static Friction Force (F_s):**
- The maximum static friction force can be calculated using the coefficient of static friction.
\[ F_s = \mu_s \cdot N = 0.5 \cdot 600 \, \text{N} = 300 \, \text{N} \]

4. **Determine the Kinetic Friction Force (F_k):**
- Once the object starts moving, the friction force acting against the motion is the kinetic friction force.
\[ F_k = \mu_k \cdot N = 0.4 \cdot 600 \, \text{N} = 240 \, \text{N} \]

5. **Calculate the Net Force (F_net):**
- The net force acting on the object when it is in motion is the applied force minus the kinetic friction force.
\[ F_{\text{net}} = F_{\text{applied}} - F_k = 300 \, \text{N} - 240 \, \text{N} = 60 \, \text{N} \]

6. **Calculate the Acceleration (a):**
- Using Newton's second law, we can find the acceleration of the object.
\[ F_{\text{net}} = m \cdot a \implies a = \frac{F_{\text{net}}}{m} = \frac{60 \, \text{N}}{60 \, \text{kg}} = 1 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of the body is \(1 \, \text{m/s}^2\).