Answer :
To solve this problem, we need to determine the car's initial velocity just before the brakes were applied. Here's a step-by-step solution:
1. Identify the given variables:
- The coefficient of kinetic friction (\( \mu \)) between the tires and the road, \( \mu = 0.7 \).
- The stopping distance (\( d \)) of the car, \( d = 120 \) meters.
- The acceleration due to gravity (\( g \)), \( g = 9.8 \) meters per second squared.
2. Understanding the physical principles involved:
We use the work-energy principle, which states that the work done by the frictional force is equal to the initial kinetic energy of the car. In other words, the frictional force acting over the stopping distance will dissipate the car’s kinetic energy.
The work done by the frictional force is:
[tex]\[ W = \text{frictional force} \times \text{distance} \][/tex]
The initial kinetic energy of the car is:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
3. Determining the frictional force:
The frictional force (\( f \)) is given by:
[tex]\[ f = \mu \times \text{normal force} \][/tex]
The normal force (in this case) is equal to the weight of the car, which is \( m \times g \). Therefore:
[tex]\[ f = \mu \times m \times g \][/tex]
4. Setting up the equation using the work-energy principle:
The work done by the frictional force must equal the car's initial kinetic energy:
[tex]\[ \mu \times m \times g \times d = \frac{1}{2} m v^2 \][/tex]
Notice that the mass (\( m \)) of the car cancels out from both sides of the equation:
[tex]\[ \mu \times g \times d = \frac{1}{2} v^2 \][/tex]
5. Solving for the initial velocity (\( v \)):
[tex]\[ v^2 = 2 \times \mu \times g \times d \][/tex]
[tex]\[ v^2 = 2 \times 0.7 \times 9.8 \times 120 \][/tex]
[tex]\[ v^2 = 1646.4 \][/tex]
Taking the square root of both sides to find \( v \):
[tex]\[ v = \sqrt{1646.4} \][/tex]
[tex]\[ v \approx 40.58 \, \text{m/s} \][/tex]
So, the car's velocity just before the brakes were applied was approximately 40.58 meters per second.
1. Identify the given variables:
- The coefficient of kinetic friction (\( \mu \)) between the tires and the road, \( \mu = 0.7 \).
- The stopping distance (\( d \)) of the car, \( d = 120 \) meters.
- The acceleration due to gravity (\( g \)), \( g = 9.8 \) meters per second squared.
2. Understanding the physical principles involved:
We use the work-energy principle, which states that the work done by the frictional force is equal to the initial kinetic energy of the car. In other words, the frictional force acting over the stopping distance will dissipate the car’s kinetic energy.
The work done by the frictional force is:
[tex]\[ W = \text{frictional force} \times \text{distance} \][/tex]
The initial kinetic energy of the car is:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
3. Determining the frictional force:
The frictional force (\( f \)) is given by:
[tex]\[ f = \mu \times \text{normal force} \][/tex]
The normal force (in this case) is equal to the weight of the car, which is \( m \times g \). Therefore:
[tex]\[ f = \mu \times m \times g \][/tex]
4. Setting up the equation using the work-energy principle:
The work done by the frictional force must equal the car's initial kinetic energy:
[tex]\[ \mu \times m \times g \times d = \frac{1}{2} m v^2 \][/tex]
Notice that the mass (\( m \)) of the car cancels out from both sides of the equation:
[tex]\[ \mu \times g \times d = \frac{1}{2} v^2 \][/tex]
5. Solving for the initial velocity (\( v \)):
[tex]\[ v^2 = 2 \times \mu \times g \times d \][/tex]
[tex]\[ v^2 = 2 \times 0.7 \times 9.8 \times 120 \][/tex]
[tex]\[ v^2 = 1646.4 \][/tex]
Taking the square root of both sides to find \( v \):
[tex]\[ v = \sqrt{1646.4} \][/tex]
[tex]\[ v \approx 40.58 \, \text{m/s} \][/tex]
So, the car's velocity just before the brakes were applied was approximately 40.58 meters per second.
