Question

73.
A car is moving with a speed of 15 m/s on a straight road is brought to rest while covering a
distance of 20/3km. If mass of car is 600 kg. Find the retarding force on the car.
(A) 200 N
(B) 225 N
(C) 450 N
(D) none of these​

Answer 1

Answer:

None of the given option is correct

Step-by-step explanation:

Mass of the car m =600 kg

We have given speed of the car u = 15 m/sec

As the car stops finally so final velocity v = 0 m/sec

Distance is given as $s=\frac{20}{3}km=\frac{20}{3}\times 1000=6666.666m$

From third equation equation of motion we know that

$v^2=u^2+2as$

$0=225+2\times a\times 6666.666$

$a=-0.01687m/sec^2$

According to newton's law force is given by

F = ma

So force will be $F=-0.01687\times 600=10.122N$

So none of the given option is correct

Answer 2

Answer:

(D)None of these

Step-by-step explanation:

We are given that

Initial speed=$u=15 m/s$

Final speed of car=$v=0$

Distance covered by car=$\frac{20}{3}km=\frac{20}{3}\times 1000=666.67 m$

$1km=1000m$

Mass of car=600 kg

We know that third equation of motion

$v^2-u^2=2aS$

Substitute the values then we get

$0-(15)^3=2\times a\times 6666.67$

$a=\frac{-225}{2\times 6666.67}=-0.01687 m/s^2$

We know that

$F=ma$

$F=600\times (-0.01687)=-10.122 N$

Hence,the retarding force on the car=-10.122 N.

Answer: (D)None of these.