Question

A motorcar is moving with a velocity of 108 km/h and it
takes 4 s to stop after the brakes are applied. Calculate the
force exerted by the brakes on the motorcar​

Answer 1

Correct Question:

A motorcar is moving with a velocity of 108 km/h 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.

Given:

• Mass of the motor car along with the passengers (M) = 1000 kg
• Time taken (t) = 4 s
• Initial velocity (u) = 108 km/h

To find:

• The exerted force by the brakes on the motor car.

Solution:

At first , convert initial velocity (u) in m/s.

$\implies\sf{u=108\times\dfrac{5}{18}\:m/s}$

$\implies\sf{u=30\:m/s}$

• Final velocity (v) = 0 m/s

Now find the exerted force (F) by the brakes on the motor car.

We know that,

${\boxed{\sf{F=\dfrac{m(v-u)}{t}}}}$

• [ Put values]

$\implies\sf{F=\dfrac{1000(0-30)}{4}}$

$\implies\sf{F=\dfrac{1000\times\:-30}{4}}$

$\implies\sf{F=-7500}$

★ Opposing force is denoted by negative sign.

Therefore, the exerted force by the brakes on the motor car is 7500 N.

Answer 2

$\underline{\underline{\blue{\bf Complete\: Question:- }}}$

$\tt A \:motorcar\: is\: moving\: with \:a\:\\\tt \:velocity\: of\: 108 km/h \:and \:it \:takes\: 4 s\: to\\\tt:stop \:after\: the\: brakes\: are\: applied.\\\tt Calculate\: the\: force \:exerted \:by\: the \:brakes\\\tt on \:the \;motorcar\: if \:it's \:mass \:is 500 kg .$

$\underline{\underline{\blue{\bf Answer:- }}}$

$\green{\tt \therefore The \:force\: applied\:by\: brakes\:is\:(-3750) Newton .}$

$\underline{\underline{\orange{\bf Step-by-step \:Explanation:- }}}$

$\underline{\green{\bf Given : }}$

$\tt:\implies A\:motorcar\:is\: moving\:with\:108km/hr. \\\tt:\implies It\:takes\:4\:s\:to \:stop\:on\: applying\:breaks.\\\tt:\implies Mass\:of\:car\:is\:500kg .$

$\underline{\red{\bf To\:Find : }}$

$\tt:\implies The\:force\: applied\:to\:stop\:the\:car .$

• By NLM second law ,

$\tt Rate\:of\:change\:of\: momentum\:is\\\tt directly \:proportional \:to \:the\: applied\: force\:\\\tt in\: the\: dir^{n}\: of\: force ..$

$\boxed{\pink{\bf \red{\bigstar} Force = \dfrac{\Delta p }{t}}}$

$\tt : \implies Force = \dfrac {\Delta p }{t}\\\\\tt:\implies Force = \dfrac{m(v-u)}{t}\\\\\tt:\implies Force = 500kg\bigg\lgroup \dfrac{ 0- 108 km/hr}{4}\bigg\rgroup\\\\\tt:\implies Force = 500kg\bigg\lgroup\dfrac{-\dfrac{\cancel{108}\times 5}{\cancel{18}}}{4}\bigg\rgroup\\\\\tt:\implies Force =\dfrac{ 500 \times -30}{4} \\\\\underline{\boxed{\red{\tt\longmapsto Force = -3750 Newtons }}}$