A 10-kg block is placed on a horizontal surface. The coefficient of friction between them is `0.2`. A horizontal force P = 15 N first acts on it in the eastward direction. Later, in addition to P a second horizontal force Q = 20 N acts on it in the northward direction.

Options

A. The block will not move when only P acts, but will move when P and Q act.
B. If the block moves, its acceleration will be `0.5 m//s^(2)`.
C. When the block moves, its direction of motion will be `tan^(-1)(4//3)` east of north.
D. When both P and Q act, the direction of the force of friction acting on the block will be `tan^(-1)(3//4)` west of south.

Answer

To solve the problem step-by-step, we will analyze the forces acting on the block and determine whether it moves or not, as well as calculate its acceleration if it does move.

### Step 1: Identify the forces acting on the block
- The block has a mass \( m = 10 \, \text{kg} \).
- The coefficient of friction \( \mu = 0.2 \).
- A horizontal force \( P = 15 \, \text{N} \) acts eastward.
- A second horizontal force \( Q = 20 \, \text{N} \) acts northward.

### Step 2: Calculate the weight of the block
The weight \( W \) of the block is given by:
\[ W = m \cdot g \]
where \( g \approx 9.8 \, \text{m/s}^2 \).
\[ W = 10 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 98 \, \text{N} \]

### Step 3: Calculate the maximum static friction
The maximum static friction \( f_{\text{max}} \) can be calculated using:
\[ f_{\text{max}} = \mu \cdot W \]
\[ f_{\text{max}} = 0.2 \cdot 98 \, \text{N} = 19.6 \, \text{N} \]

### Step 4: Analyze the forces when only force \( P \) is acting
When only force \( P \) is acting:
- The applied force \( P = 15 \, \text{N} \) (eastward).
- The maximum static friction \( f_{\text{max}} = 19.6 \, \text{N} \).

Since \( P < f_{\text{max}} \), the block does not move when only force \( P \) is applied.

### Step 5: Analyze the forces when both forces \( P \) and \( Q \) are acting
Now, consider both forces \( P \) and \( Q \):
- The resultant force \( R \) can be calculated using the Pythagorean theorem since they are perpendicular:
\[ R = \sqrt{P^2 + Q^2} = \sqrt{(15 \, \text{N})^2 + (20 \, \text{N})^2} = \sqrt{225 + 400} = \sqrt{625} = 25 \, \text{N} \]

### Step 6: Determine if the block moves with both forces
Now compare the resultant force \( R \) with the maximum static friction:
- Resultant force \( R = 25 \, \text{N} \)
- Maximum static friction \( f_{\text{max}} = 19.6 \, \text{N} \)

Since \( R > f_{\text{max}} \), the block will move when both forces are applied.

### Step 7: Calculate the acceleration of the block
Using Newton's second law:
\[ F_{\text{net}} = m \cdot a \]
The net force acting on the block when it moves is:
\[ F_{\text{net}} = R - f_{\text{max}} = 25 \, \text{N} - 19.6 \, \text{N} = 5.4 \, \text{N} \]
Now, we can find the acceleration \( a \):
\[ 5.4 \, \text{N} = 10 \, \text{kg} \cdot a \implies a = \frac{5.4 \, \text{N}}{10 \, \text{kg}} = 0.54 \, \text{m/s}^2 \]

### Step 8: Determine the direction of motion
The direction of motion can be determined using the angle \( \theta \):
\[ \tan(\theta) = \frac{Q}{P} = \frac{20}{15} = \frac{4}{3} \]
Thus, the angle \( \theta \) can be found using:
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

### Summary of Results
1. The block does not move when only force \( P \) is applied.
2. The block moves when both forces \( P \) and \( Q \) are applied.
3. The acceleration of the block is \( 0.54 \, \text{m/s}^2 \).
4. The direction of motion is given by \( \tan^{-1}(4/3) \).