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 step by step, we will follow these calculations:

### Step 1: Calculate the Normal Force
The normal force (N) acting on the body is equal to the weight of the body when it is on a horizontal surface.

\[ N = m \cdot g \]

Where:
- \( m = 60 \, \text{kg} \) (mass of the body)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

\[ N = 60 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 588 \, \text{N} \]

### Step 2: Calculate the Static Friction Force
The force of static friction (Fs) can be calculated using the coefficient of static friction (μs) and the normal force (N).

\[ F_s = \mu_s \cdot N \]

Where:
- \( \mu_s = 0.5 \)

\[ F_s = 0.5 \cdot 588 \, \text{N} = 294 \, \text{N} \]

### Step 3: Calculate the Kinetic Friction Force
Once the body starts moving, the force of kinetic friction (Fk) acts on it. This can be calculated using the coefficient of kinetic friction (μk) and the normal force (N).

\[ F_k = \mu_k \cdot N \]

Where:
- \( \mu_k = 0.4 \)

\[ F_k = 0.4 \cdot 588 \, \text{N} = 235.2 \, \text{N} \]

### Step 4: Calculate the Net Force Acting on the Body
When the body is pushed, the applied force (F) is equal to the static friction force when it starts moving. After it starts moving, the same force continues to act, but now we consider the kinetic friction.

\[ F_{\text{net}} = F - F_k \]

Where:
- \( F = F_s = 294 \, \text{N} \)

\[ F_{\text{net}} = 294 \, \text{N} - 235.2 \, \text{N} = 58.8 \, \text{N} \]

### Step 5: Calculate the Acceleration
Using Newton's second law, we can find the acceleration (a) of the body.

\[ F_{\text{net}} = m \cdot a \]

Rearranging gives:

\[ a = \frac{F_{\text{net}}}{m} \]

Substituting the values:

\[ a = \frac{58.8 \, \text{N}}{60 \, \text{kg}} = 0.98 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the body is approximately \( 0.98 \, \text{m/s}^2 \).

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