A 15 kg mass is accelerated from rest with a force of 100 N. As it moves faster, friction and air resistance create an oppositively directed retarding force given by FR = A + Bv, where A = 25 N and B = 0.5 N/m/s. At what velocity does the acceleration equal to one half of the initial acceleration?

Options

A. `25ms^(-1)`
B. `50m//s`
C. `75 m//s`
D. `100m//s`

Answer

To solve the problem step by step, we will follow the process of calculating the initial acceleration, then determining the velocity at which the acceleration is half of the initial acceleration.

### Step 1: Calculate the Initial Acceleration
The initial acceleration can be calculated using Newton's second law of motion. The net force acting on the mass is the applied force minus the retarding force.

1. **Given:**
- Mass (m) = 15 kg
- Applied force (F) = 100 N
- Retarding force (FR) at rest = A = 25 N (since v = 0)

2. **Calculate the net force:**
\[ F_{\text{net}} = F - F_R = 100\, \text{N} - 25\, \text{N} = 75\, \text{N} \]

3. **Using Newton's second law (F = ma):**
\[ a = \frac{F_{\text{net}}}{m} = \frac{75\, \text{N}}{15\, \text{kg}} = 5\, \text{m/s}^2 \]

### Step 2: Set Up the Equation for Half of the Initial Acceleration
We need to find the velocity at which the acceleration equals half of the initial acceleration.

1. **Calculate half of the initial acceleration:**
\[ \frac{a}{2} = \frac{5\, \text{m/s}^2}{2} = 2.5\, \text{m/s}^2 \]

2. **Set up the equation for acceleration when the object is moving with velocity \( v \):**
The retarding force now includes both A and the term \( Bv \):
\[ F_R = A + Bv = 25\, \text{N} + 0.5\, \text{N/m/s} \cdot v \]

3. **Using Newton's second law again:**
\[ \frac{F - F_R}{m} = \frac{a}{2} \]
Substituting the values:
\[ \frac{100\, \text{N} - (25\, \text{N} + 0.5\, \text{N/m/s} \cdot v)}{15\, \text{kg}} = 2.5\, \text{m/s}^2 \]

### Step 3: Solve for Velocity \( v \)
1. **Rearranging the equation:**
\[ 100 - 25 - 0.5v = 15 \cdot 2.5 \]
\[ 75 - 0.5v = 37.5 \]

2. **Isolate \( v \):**
\[ 75 - 37.5 = 0.5v \]
\[ 37.5 = 0.5v \]
\[ v = \frac{37.5}{0.5} = 75\, \text{m/s} \]

### Final Answer
The velocity at which the acceleration equals one half of the initial acceleration is \( v = 75\, \text{m/s} \).

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