A toy car can deliver a constant power of `20W`. The resistive force on the car is `alphav`, where `v` is velocity of car in `m//s`. If maximum velocity of car is `2m//s`, the value of `alpha` is .

Options

A. `10 kg//s`
B. `5kg//s`
C. `15kg//s`
D. `5 kg//s^(2)`

Explanation

To solve the problem, we need to find the value of \(\alpha\) given the constant power of the toy car and the resistive force acting on it. Here’s the step-by-step solution:

### Step 1: Understand the relationship between power, force, and velocity
The power \(P\) delivered by the car can be expressed as:
\[ P = F \cdot v \]
where \(F\) is the net force acting on the car and \(v\) is the velocity of the car.

### Step 2: Identify the resistive force
According to the problem, the resistive force \(F\) acting on the car is given by:
\[ F = \alpha v \]
where \(\alpha\) is a constant we need to find.

### Step 3: Substitute the resistive force into the power equation
At maximum velocity, the power can be expressed as:
\[ P = \alpha v \cdot v = \alpha v^2 \]
This means that the power delivered by the car is equal to the resistive force multiplied by the velocity.

### Step 4: Plug in the known values
We know that the maximum power \(P\) is \(20 \, \text{W}\) and the maximum velocity \(v\) is \(2 \, \text{m/s}\). Substituting these values into the equation gives:
\[ 20 = \alpha (2^2) \]
\[ 20 = \alpha \cdot 4 \]

### Step 5: Solve for \(\alpha\)
Now, we can solve for \(\alpha\):
\[ \alpha = \frac{20}{4} = 5 \, \text{kg/s} \]

### Conclusion
Thus, the value of \(\alpha\) is \(5 \, \text{kg/s}\).

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