Question

6. A car has a mass of 2 tonnes (1 tonne = 1000 kg) and decelerates at 2.8 m/s². Calculate the force being applied to the brakes.

Use the formula:
force = mass x acceleration​

Answer 1

F_B = -5600 N

Step-by-step explanation:

We are tasked with determining the force applied to the brakes of a car with a mass of 2 tons that decelerates at a rate of 2.8 m/s². We will use Newton's second law of motion to calculate this force.

`\hrulefill`

Start by constructing a free-body diagram to visualize the forces acting on the car. I have attached this image for you to view.

We can neglect the forces acting in the vertical direction and focus on the one acting in the horizontal. Using the following formula:

`\sum \vec F_x=m \vec a \ \Big[=m\vec a \ \because \text{there is } \vec a\Big]`

Let's add up the forces:

`\Longrightarrow \vec F_B\cos(180 \textdegree)=m\vec a\\\\\\\\\Longrightarrow -\vec F_B=m\vec a\\\\\\\\\therefore \vec F_B=-m\vec a`

Using the above equation, where:

• 'F_B' is the force applied to the brakes (in Newtons, N)
• 'm' is the mass of the car (kg)
• 'a' is the acceleration of the car (m/s²)

Substitute all we know into the equation and solve for 'F_B':

`\Longrightarrow \vec F_B=-\Big(2 \text{ tons}\cdot \frac{1000 \text{ kg}}{1 \text{ ton}} \Big)(2.8 \text{ m/s}^2)\\\\\\\\\Longrightarrow \vec F_B=-(2000 \text{ kg})(2.8\text{ m/s}^2)\\\\\\\\\therefore \boxed{\vec F_B=-5600 \ N}`

Therefore, the force being applied to the brakes is -5,600 Newtons. The negative sign indicates the force direction opposing the car's motion.