A `1.00xx10^(2)` kg crate is being pushed across a horizontal floor by a force `vec(P)` that makes an angle of `30.0^(@)` below the horizontal. The coefficient of kinetic friction is 0.200. What should be the magnitude of `vec(P)`, so that the net work done by it and the kinetic frictional force is zero ?

Options

A. 256N
B. 354N
C. 203N
D. 287N

Answer

To solve the problem, we need to find the magnitude of the force \( \vec{P} \) such that the net work done by it and the kinetic frictional force is zero. Here’s a step-by-step solution:

### Step 1: Identify the forces acting on the crate
1. The gravitational force \( \vec{W} \) acting downwards, which is equal to \( mg \) where \( m = 100 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
2. The applied force \( \vec{P} \) at an angle of \( 30^\circ \) below the horizontal.
3. The normal force \( \vec{N} \) acting upwards.
4. The frictional force \( \vec{F}_{\text{friction}} \) acting opposite to the direction of motion.

### Step 2: Calculate the weight of the crate
\[ W = mg = 100 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 980 \, \text{N} \]

### Step 3: Resolve the applied force \( \vec{P} \) into components
- The horizontal component of \( \vec{P} \):
\[ P_x = P \cos(30^\circ) = P \cdot \frac{\sqrt{3}}{2} \]
- The vertical component of \( \vec{P} \):
\[ P_y = P \sin(30^\circ) = P \cdot \frac{1}{2} \]

### Step 4: Determine the normal force \( \vec{N} \)
The normal force can be expressed as:
\[ N = W - P_y = mg - P \cdot \frac{1}{2} \]
Substituting the weight:
\[ N = 980 \, \text{N} - P \cdot \frac{1}{2} \]

### Step 5: Calculate the frictional force
The frictional force \( F_{\text{friction}} \) is given by:
\[ F_{\text{friction}} = \mu N = 0.2 \left( 980 - \frac{P}{2} \right) \]

### Step 6: Set up the equation for zero net work
For the net work done to be zero, the forward force must equal the frictional force:
\[ P_x = F_{\text{friction}} \]
Substituting the expressions we derived:
\[ P \cdot \frac{\sqrt{3}}{2} = 0.2 \left( 980 - \frac{P}{2} \right) \]

### Step 7: Solve for \( P \)
Expanding the equation:
\[ P \cdot \frac{\sqrt{3}}{2} = 196 - 0.1P \]
Rearranging gives:
\[ P \cdot \frac{\sqrt{3}}{2} + 0.1P = 196 \]
Factoring out \( P \):
\[ P \left( \frac{\sqrt{3}}{2} + 0.1 \right) = 196 \]
Calculating the left side:
\[ P \left( \frac{\sqrt{3}}{2} + 0.1 \right) = P \left( 0.866 + 0.1 \right) = P \cdot 0.966 \]
Thus:
\[ P = \frac{196}{0.966} \approx 203.5 \, \text{N} \]

### Step 8: Final calculation
To find the exact value:
\[ P \approx 203.5 \, \text{N} \]

### Summary
The magnitude of the force \( \vec{P} \) required to ensure that the net work done by it and the kinetic frictional force is zero is approximately **203.5 N**.

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