A 500 kg horse pulls a cart of mass 1500 kg along a level road with an acceleration of `1ms^(-2)`. ms . If the coefficient of sliding friction is 0.2, then the force exerted by the horse in forward direction is

Options

A. 3000 N
B. 4000 N
C. 5000 N
D. 6000 N

Answer

To solve the problem, we need to find the force exerted by the horse in the forward direction. We will follow these steps:

### Step 1: Identify the given data
- Mass of the horse (m_horse) = 500 kg
- Mass of the cart (m_cart) = 1500 kg
- Total mass (M) = m_horse + m_cart = 500 kg + 1500 kg = 2000 kg
- Acceleration (a) = 1 m/s²
- Coefficient of sliding friction (μ) = 0.2
- Acceleration due to gravity (g) = 9.81 m/s² (approximately 10 m/s² for simplicity)

### Step 2: Calculate the normal force (N)
The normal force acting on the system (both horse and cart) is equal to the total weight of the system:
\[ N = M \cdot g = 2000 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20000 \, \text{N} \]

### Step 3: Calculate the frictional force (F_friction)
The frictional force can be calculated using the formula:
\[ F_{\text{friction}} = \mu \cdot N \]
Substituting the values:
\[ F_{\text{friction}} = 0.2 \cdot 20000 \, \text{N} = 4000 \, \text{N} \]

### Step 4: Apply Newton's second law
According to Newton's second law, the net force (F_net) acting on the system is given by:
\[ F_{\text{net}} = F - F_{\text{friction}} \]
Where \( F \) is the force exerted by the horse. The net force is also equal to the total mass multiplied by the acceleration:
\[ F_{\text{net}} = M \cdot a = 2000 \, \text{kg} \cdot 1 \, \text{m/s}^2 = 2000 \, \text{N} \]

### Step 5: Set up the equation
Setting the two expressions for net force equal gives us:
\[ F - F_{\text{friction}} = M \cdot a \]
Substituting the values we have:
\[ F - 4000 \, \text{N} = 2000 \, \text{N} \]

### Step 6: Solve for F
Rearranging the equation to solve for \( F \):
\[ F = 2000 \, \text{N} + 4000 \, \text{N} \]
\[ F = 6000 \, \text{N} \]

### Final Answer
The force exerted by the horse in the forward direction is **6000 N**.

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