Question

You are driving your 1500 kg car at 20 m/s down a hill with a 5.0° slope when a deer suddenly jumps out onto the roadway. You slam on your brakes, skidding to a stop. How far do you skid before stopping if the kinetic friction force between your tires and the road is 1.2 * 104 N? Solve this problem using conservation of energy.

Answer 1

To solve this problem using conservation of energy, calculate the initial kinetic energy of the car and the work done by the friction force to bring the car to a stop. Set the work done equal to the negative change in kinetic energy and solve for the distance.

Step-by-step explanation:

To solve this problem using conservation of energy, we can start by calculating the initial kinetic energy of the car. The kinetic energy is given by the formula KE = (1/2) * mass * velocity^2. Plugging in the values, we have KE = (1/2) * 1500 kg * (20 m/s)^2 = 300,000 J.

Next, we need to calculate the work done by the friction force to bring the car to a stop. The work done is given by the formula work = force * distance. Plugging in the values, we have work = (1.2 * 10^4 N) * distance. Since the car comes to a stop, the work done by the friction force is equal to the negative change in kinetic energy: -300,000 J.

Simplifying the equation, we have (1.2 * 10^4 N) * distance = -300,000 J. Solving for distance, we find that distance = -300,000 J / (1.2 * 10^4 N) = -25 meters.

Since distance cannot be negative, we take the absolute value of the distance, giving us a skid distance of 25 meters.

Answer 2

The car skids a distance of
`\(25 \, \text{m}\)` before coming to a stop.

Using the conservation of energy, we can find the distance the car skids before stopping.

The initial kinetic energy
`(\(KE_i\))` of the car is converted into work done against friction to bring the car to a stop. The work done against friction will be equal to the initial kinetic energy.

The initial kinetic energy of the car
`(\(KE_i\))` is given by:

`\[KE_i = (1)/(2)mv^2\]`

where:

m = mass of the car = 1500 kg

v = initial velocity = 20 m/s

`\[KE_i = (1)/(2) * 1500 \, \text{kg} * (20 \, \text{m/s})^2\]`

`\[KE_i = (1)/(2) * 1500 \, \text{kg} * 400 \, \text{m}^2/\text{s}^2\]`

`\[KE_i = 300,000 \, \text{J}\]`

This initial kinetic energy is equal to the work done against friction:

`\[Work = KE_i\]`

The work done against friction (Work) is given by the force of friction multiplied by the distance (d):

`\[Work = \text{Friction force} * d\]`

`\[300,000 \, \text{J} = (1.2 * 10^4 \, \text{N}) * d\]`

Solving for d:

`\[d = \frac{300,000 \, \text{J}}{1.2 * 10^4 \, \text{N}}\]`

`\[d = 25 \, \text{m}\]`