A motor car of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is reduced to 18 km/h in 4s by an unbalanced external force. The magnitude of force applied is

320 N

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4800 N

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6000 N

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2400 N

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Solution

The correct option is C $6000 N$
Given,
$m = 1200 k g$
$u = 90 k m / h = \frac{90 \times 1000}{60 \times 60} = 25 m / s$

$v = 18 k m / h = \frac{18 \times 1000}{60 \times 60} = 5 m / s$
$t = 4 s e c$

1st equation of motion,
$v = u + a t$
$5 = 25 + 4 a$
$a = - \frac{20}{4} = - 5 m / s^{2}$

Force, $F = m a = - 1200 \times 5 = - 6000 N$
$| F | = 6000 N$
The correct option is C.

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Similar questions

Q. A motorcar of mass

1200 k g

is moving along a straight line with a uniform velocity of

90 k m / h

. Its velocity is slowed down to

18 k m / h

4 s

by an unbalanced external force. Calculate the magnitude of the force required.

A car of mass 1200 kg is moving along a straight line with a uniform velocity of 90 $\frac{k m}{h}$ . Its velocity is slowed down to 18 ${k m h}^{- 1}$ in 4 s by an unbalanced external force. Calculate the change in momentum and the magnitude of the force required.

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A motor car of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is reduced to 18 km/h in 4s by an unbalanced external force. The magnitude of force applied is

The correct option is C 6000NGiven,m=1200kgu=90km/h=90×100060×60=25m/sv=18km/h=18×100060×60=5m/st=4sec1st equation of motion,v=u+at5=25+4aa=−204=−5m/s2Force, F=ma=−1200×5=−6000N|F|=6000NThe correct option is C.