Question

motorcar is moving with 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 .

Answer 1

Good morning!
According Newton's Second Law of Motion,
$f = ma$
(as long as mass remains constant)
Here we have,
$u = \frac{108 \times 1000}{1 \times 3600} \: metre \: per \: second.$
$v = 0 \: m {s}^{ - 1}$
$t = 4s$
$m = 1000kg$
we know
$f = ma$
and
$a = \frac{v - u}{t}$
Thus
$f = m \: \times \frac{v - u}{t}$
$f = - \frac{30}{4} \times 1000 \\ = - 7500 \: newton$
Thus it is a retarding force.
Hope it helps!

Answer 2

${ \huge{ \bold{ \underline{ \underline{ \red{Question:-}}}}}}$

A Motorcar is moving with 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 .

_______________

${ \huge{ \bold{ \underline{ \underline{ \orange{Answer:-}}}}}}$

✒Given : -

• Initial Velocity(u) = 108 km/h = 30m/s-¹
• Final Velocity(v) = 0m/s-¹
• Total Mass(m) = 1000kg
• Time Taken(t) = 4 sec.

✒To Find : -

• The force exerted by the brakes on the motorcar = ?

✒Formula Used : -

$\leadsto\sf{{ \small{ \bold{ \bold{ \bold{ \green{Force=\dfrac{m(v-u)}{t}}}}}}}}$

➦On Substituting Values : -

$\dashrightarrow\sf{F=\dfrac{1000kg\times{(0-30)}\:{m/s}^{-1}}{4s}}$

$\leadsto\sf{{ \large{ \boxed{ \bold{ \bold{ \blue{-7500\:N}}}}}}}$

☛ The (-) sign tells us about that the force exerted by the brakes is opposite to the direction of motion of the motorcar ..