Stopping Distances
In this lesson we learn how to calculate braking force.
IN THIS LESSON:
Calculating braking force
Braking force definition
Braking force equation
How to use braking force equation
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Practice Questions
Question 1: Define the term 'work done' and state its unit.
Answer: Work done is the energy transferred when a force moves an object. Its unit is the Joule (J).
Question 2: Define 'kinetic energy' and state its unit.
Answer: Kinetic energy is the energy of a moving object. Its unit is the Joule (J).
Question 3: Describe the energy transfer that takes place when a car brakes.
Answer: When a car brakes, its kinetic energy is transferred by the brakes into thermal energy (heat) in the brake pads and brake discs, and also into sound energy.
Question 4: A car of mass 1200 kg is travelling at a velocity of 10 m/s. Calculate its kinetic energy.
Answer Walkthrough:
- Given: Mass (m) = 1200 kg, Velocity (v) = 10 m/s.
- Formula: $KE = 0.5 \times m \times v^2$
- Calculation: $KE = 0.5 \times 1200 \text{ kg} \times (10 \text{ m/s})^2 = 600 \times 100 = 60,000 \text{ J}$.
Answer: The kinetic energy is 60,000 J.
Question 5: A lorry of mass 8000 kg travels at a velocity of 20 m/s. Calculate its kinetic energy.
Answer Walkthrough:
- Given: Mass (m) = 8000 kg, Velocity (v) = 20 m/s.
- Formula: $KE = 0.5 \times m \times v^2$
- Calculation: $KE = 0.5 \times 8000 \text{ kg} \times (20 \text{ m/s})^2 = 4000 \times 400 = 1,600,000 \text{ J}$.
Answer: The kinetic energy is 1,600,000 J.
Question 6: How does doubling a car's velocity affect its kinetic energy? Justify your answer using the formula.
Answer: Doubling the velocity quadruples the kinetic energy. This is because velocity is squared in the kinetic energy formula ($KE = 0.5 \times m \times v^2$). So, if you replace v with 2v, the kinetic energy becomes $0.5 \times m \times (2v)^2 = 0.5 \times m \times 4v^2 = 4 \times (0.5 \times m \times v^2)$.
Question 7: A braking force of 5000 N is applied to a car over a distance of 20 m. Calculate the work done by the brakes.
Answer Walkthrough:
- Given: Force (F) = 5000 N, Distance (d) = 20 m.
- Formula: $W = F \times d$
- Calculation: $W = 5000 \text{ N} \times 20 \text{ m} = 100,000 \text{ J}$.
Answer: The work done by the brakes is 100,000 J.
Question 8: A car stops over a braking distance of 40 m. The average braking force is 8000 N. What was the initial kinetic energy of the car?
Answer Walkthrough:
According to the work-energy principle, the initial kinetic energy is equal to the work done.
- Given: Force (F) = 8000 N, Distance (d) = 40 m.
- Formula: $KE = W = F \times d$
- Calculation: $KE = 8000 \text{ N} \times 40 \text{ m} = 320,000 \text{ J}$.
Answer: The initial kinetic energy of the car was 320,000 J.
Question 9: A braking force does 120,000 J of work over a distance of 30 m. Calculate the size of the braking force.
Answer Walkthrough:
- Given: Work Done (W) = 120,000 J, Distance (d) = 30 m.
- Formula: $W = F \times d$, rearrange to $F = W/d$.
- Calculation: $F = 120,000 \text{ J} / 30 \text{ m} = 4000 \text{ N}$.
Answer: The braking force is 4000 N.
Question 10: A car has 450,000 J of kinetic energy. It comes to a stop with a braking distance of 50 m. Calculate the average braking force.
Answer Walkthrough:
- Given: Kinetic Energy (KE) = 450,000 J, Distance (d) = 50 m.
- Principle: $W = KE$, so $F \times d = KE$.
- Formula: $F = KE/d$
- Calculation: $F = 450,000 \text{ J} / 50 \text{ m} = 9000 \text{ N}$.
Answer: The average braking force is 9000 N.
Question 11: A motorcycle of mass 250 kg is travelling at 30 m/s and stops in a braking distance of 45 m. Calculate the average braking force.
Answer Walkthrough:
Step 1: Calculate the initial kinetic energy.
- $KE = 0.5 \times m \times v^2 = 0.5 \times 250 \text{ kg} \times (30 \text{ m/s})^2 = 125 \times 900 = 112,500 \text{ J}$.
Step 2: Use the work-energy principle to find the force.
- $W = KE$, so $F \times d = 112,500 \text{ J}$.
- Formula: $F = W/d$
- Calculation: $F = 112,500 \text{ J} / 45 \text{ m} = 2500 \text{ N}$.
Answer: The average braking force is 2500 N.
Question 12: A car (mass 1000 kg) and a van (mass 2000 kg) are both travelling at 25 m/s. If they both stop over the same braking distance, compare the average braking forces required.
Answer Walkthrough:
Step 1: Calculate the KE for each vehicle.
- $KE_{car} = 0.5 \times 1000 \text{ kg} \times (25 \text{ m/s})^2 = 500 \times 625 = 312,500 \text{ J}$.
- $KE_{van} = 0.5 \times 2000 \text{ kg} \times (25 \text{ m/s})^2 = 1000 \times 625 = 625,000 \text{ J}$.
Step 2: Compare the braking forces.
Since $F = KE/d$ and the distance (d) is the same for both, the force is proportional to the kinetic energy.
The van has twice the kinetic energy of the car, so it will require twice the braking force to stop over the same distance.
Answer: The braking force required for the van is twice that of the car because its kinetic energy is twice as large.
Question 13: A car's brakes can provide a maximum force of 8000 N. If the car has 400,000 J of kinetic energy, what is the minimum braking distance?
Answer Walkthrough:
- Given: Work Done (W) = 400,000 J, Force (F) = 8000 N.
- Principle: $W = KE$, so $F \times d = KE$.
- Formula: $d = KE/F$
- Calculation: $d = 400,000 \text{ J} / 8000 \text{ N} = 50 \text{ m}$.
Answer: The minimum braking distance is 50 m.
Question 14: A car of mass 1400 kg is travelling at 20 m/s. It brakes to a stop over a distance of 40 m.
a) Calculate the total kinetic energy that must be transferred.
b) Calculate the average braking force required.
Answer Walkthrough:
a)
- Step 1: Calculate the initial kinetic energy.
- $KE = 0.5 \times m \times v^2 = 0.5 \times 1400 \text{ kg} \times (20 \text{ m/s})^2 = 700 \times 400 = 280,000 \text{ J}$.
b)
- Step 2: Use the work-energy principle to find the force.
- $F = KE/d$
- $F = 280,000 \text{ J} / 40 \text{ m} = 7000 \text{ N}$.
Answer:
a) The kinetic energy is 280,000 J.
b) The average braking force is 7000 N.
Question 15: A driver's car has a maximum braking force of 7000 N on a dry road. If the driver is travelling at 30 mph (13.4 m/s) on an icy road, where the braking force is reduced to 500 N, what will be the braking distance?
Answer Walkthrough:
Step 1: Calculate the initial kinetic energy.
First, we need the car's mass. Let's assume a typical car mass of 1200 kg.
- $KE = 0.5 \times m \times v^2 = 0.5 \times 1200 \text{ kg} \times (13.4 \text{ m/s})^2 = 600 \times 179.56 = 107,736 \text{ J}$.
Step 2: Use the work-energy principle to find the distance.
The work done must equal this kinetic energy.
- Given: Force on ice (Fice) = 500 N, Work Done (W) = 107,736 J.
- Formula: $d = W/F$
- Calculation: $d = 107,736 \text{ J} / 500 \text{ N} = 215.472 \text{ m}$.
Answer: The braking distance on the icy road would be approximately 215 m, which is a huge distance.