Question

a car of mass 720 kg moving at a speed of 54 km per hour is stopped by applying brakes in 10 seconds calculate the force applied by the brakes​

Answer 1

Answer:

Given :-

• A car of mass 720 kg moving at a speed of 54 km/h is stopped by applying brakes in 10 seconds.

To Find :-

• What is the force applied by the brakes.

Formula Used :-

$\clubsuit$ First Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

$\clubsuit$ Force Formula :

$\mapsto \sf\boxed{\bold{\pink{Force =\: Mass \times Acceleration}}}$

Solution :-

First, we have to convert the initial velocity km/h into m/s :

$\implies \sf Initial\: Velocity =\: 54\: km/h$

$\implies \sf Initial\: Velocity =\: 54 \times \dfrac{5}{18}\: m/s\: \: \bigg\lgroup \sf\bold{\pink{1\: km/h =\: \dfrac{5}{18}\: m/s}}\bigg\rgroup$

$\implies \sf Initial\: Velocity =\: \dfrac{270}{18}\: m/s$

$\implies \sf\bold{\blue{Initial\: Velocity =\: 15\: m/s}}$

Now, we have to find the acceleration :

Given :

• Initial Velocity (u) = 15 m/s
• Final Velocity (v) = 0 m/s
• Time Taken (t) = 10 seconds

According to the question by using the formula we get,

$\leadsto \sf 0 =\: 15 + a(10)$

$\leadsto \sf 0 =\: 15 + 10a$

$\leadsto \sf 0 - 15 =\: 10a$

$\leadsto \sf - 15 =\: 10a$

$\leadsto \sf \dfrac{- 15}{10} =\: a$

$\leadsto \sf - 1.5 =\: a$

$\leadsto \sf\bold{\purple{a =\: - 1.5\: m/s^2}}$

Here, the negetive sign indicates retardation.

Now, we have to find the force applied by the brakes :

Given :

• Mass (m) = 720 kg
• Acceleration (a) = 1.5 m/s²

According to the question by using the formula we get,

$\longrightarrow \sf Force =\: 720 \times 1.5$

$\longrightarrow \sf\bold{\red{Force =\: 1080\: N}}$

${\small{\bold{\underline{\therefore\: The\: force\: applied\: by\: the\: brakes\: is\: 1080\: N\: .}}}}$

Answer 2

Given,

$\rm \: \to \: mass \: of \: the \: car \: \: m = 720 \: kg \\ \\ \rm \to \: initial \: velocity \: \: u = 54 \:k m {h}^{ - 1} \\ \\ \rm \implies\:u= \frac{54 \times 1000}{3600} \: m {s}^{ - 1} \\ \\ \rm \implies\:u = 15 \: m {s}^{ - 1} \\ \\ \rm \to \: final \: velocity \: \: v = 0 \: \: \{ \sf \: stopped \: \} \\ \\ \rm \to \: time \: \: t = 10 \: s$

To find :

The force (F) applied by the brakes.

Solution :

We know that,

$\: \: \: \green{ \boxed{ \bf \: F = ma}}$

To find force at first we have to find acceleration (a).

And it can be found by the following formula :

$\bf \: v = u + at \\ \\ \rm \implies\:a = \frac{v - u}{t} \\ \\ \rm \implies\:a = \frac{0 - 15}{10} \\ \\ \rm \implies\:a = - 1.5 \: m {s}^{ - 2}$

The negative sign implies that the velocity is decreasing as well as retardation.

Now,

The force applied by the brakes,

$\bf \: F = ma \\ \\ \rm \implies\: F= 720 \times ( - 1.5) \\ \\ \rm \implies\:F = - 1080 \: N$