Question

A motorcar is moving with a velocity of 108 m/s and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg

Answer 1

Provided that:

• Initial velocity = 108 m/s
• Time taken = 4 seconds
• Final velocity = 0 m/s
• Mass = 1000 kg

To calculate:

• Force exerted by the brakes on the motorcar.

Solution:

• Force exerted by the brakes on the motorcar = -27000 Newton

Using concept:

• Formula to find force

Using formula:

• F = ma

Where, F denotes force, m denotes mass of the body, a denotes acceleration.

Knowledge required:

• The SI unit of force is derived from the SI unit of mass and acceleration respectively as the SI unit of mass is kg and the SI unit of acceleration is m/s² therefore, when combined the International system (SI) unit of force becames kg m/s²

• But commonly the SI unit of force is known as Newton!

Required solution:

$:\implies \sf Force \: = Mass \times Acceleration \\ \\ :\implies \sf F \: = ma \\ \\ :\implies \sf F \: = m \times \dfrac{v-u}{t} \\ \\ :\implies \sf F \: = 1000 \times \dfrac{0-108}{4} \\ \\ :\implies \sf F \: = 1000 \times \dfrac{-108}{4} \\ \\ :\implies \sf F \: = 1000 \times -27 \\ \\ :\implies \sf F \: = -27000 \: N$

Therefore, force exerted by the brakes on the motorcar = -27000 Newton

Answer 2

Question :

A motorcar is moving with a velocity of 108 m/s and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg.

Answer :

Initial velocity (u) = 108 m/s

Final velocity (v) = 0

Time taken (t) = 4 s

Mass (m) = 1000kg

According to the first equation of motion

$\begin{gathered}\\ \: \: \: \: \bigstar\large{\red{\underline{\boxed{\bf{\pink{v= u + at}}}}}}\\\\\end{gathered}$

$\begin{gathered}\\\implies\tt 0=108+a\times{4}\\\\\end{gathered}$

$\begin{gathered}\implies\tt 0=108+4a \\\\\end{gathered}$

$\begin{gathered}\implies\tt -108 = 4a \\\\\end{gathered}$

$\begin{gathered}\implies\tt a=\cancel\dfrac{-108}{4}\\\\\end{gathered}$

$\begin{gathered}\implies\tt a= -27m/s^2\\\\\end{gathered}$

$\begin{gathered}\therefore{\tt{\pink{Acceleration\:of\:motorcar=-27m/s^2}}}\\\\\end{gathered}$

As we know

Minus shows retardation of motor car

________________________

$\begin{gathered}\\{Let's\:find\:out\:the\:force\: exerted\:by\:brakes}\\\\\end{gathered}$

$\begin{gathered}\bigstar{\boxed{\bf{Force=mass\times acceleration}}}\\\\\end{gathered}$

$\begin{gathered}\implies\tt F=m\times{a}\\\\\end{gathered}$

$\begin{gathered}\implies\tt F=1000\times{(-27)}\\\\\end{gathered}$

$\begin{gathered}\implies\tt F=-2700\:N \\\\\end{gathered}$