A 1580 kg car is traveling with a speed of 15.0 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 50.0 m ?

Options

A. 1030 N
B. 2490 N
C. 3560 N
D. 4010 N

Answer

To solve the problem of finding the magnitude of the horizontal net force required to bring a 1580 kg car traveling at 15.0 m/s to a halt in a distance of 50.0 m, we can follow these steps:

### Step 1: Identify the known values
- Mass of the car (m) = 1580 kg
- Initial velocity (u) = 15.0 m/s
- Final velocity (v) = 0 m/s (since the car comes to a halt)
- Distance (s) = 50.0 m

### Step 2: Use the third equation of motion
We can use the third equation of motion, which states:
\[ v^2 = u^2 + 2as \]
Where:
- v = final velocity
- u = initial velocity
- a = acceleration (which will be negative since it's deceleration)
- s = distance

### Step 3: Rearranging the equation to find acceleration
Rearranging the equation to solve for acceleration (a):
\[ a = \frac{v^2 - u^2}{2s} \]

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation:
\[ a = \frac{0^2 - (15.0)^2}{2 \times 50.0} \]
\[ a = \frac{0 - 225}{100} \]
\[ a = \frac{-225}{100} \]
\[ a = -2.25 \, \text{m/s}^2 \]

### Step 5: Calculate the magnitude of the horizontal net force
Using Newton's second law, the net force (F) can be calculated using the formula:
\[ F = m \cdot a \]
Since we are interested in the magnitude of the force, we will take the absolute value of acceleration:
\[ F = m \cdot |a| \]
Substituting the values:
\[ F = 1580 \, \text{kg} \cdot 2.25 \, \text{m/s}^2 \]
\[ F = 3555 \, \text{N} \]

### Step 6: Final answer
The magnitude of the horizontal net force required to bring the car to a halt is:
\[ F = 3555 \, \text{N} \]