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(a) State and explain Newton’s second law of motion.
(b) A 1000 kg vehicle moving with a speed of 20 m/s is brought to rest in a distance of 50 meters:
(i) Find the acceleration.
(ii) Calculate the unbalanced force acting on the vehicle.
(a) Newton’s second law of motion: The rate of change of momentum of a body with respect to the time is directly proportional to the applied force, and takes place in the direction in which the force acts.
Consider a body of mass m having initial velocity u.
The initial momentum of the body$=mu$
Let us assume a force $F$ acts on the body for time $t$ causing the final velocity to be $v$.
The final momentum of the body $=mv$
Now the change in momentum$=mv-mu$
And the time taken for this change is $t$
Rate of the change of the momentum $=m(\frac{v-u}{t}$
$=ma$ [because $a=\frac{v-u}{t}$]
And according to Newton's second law of motion
$F\propto ma$
Or $F=kma$
Here $k$ is a constant. In certain conditions when $m=1\ kg.$, $a=1\ m/s^2$ and $F=1\ N$
Then, $k=1$
So $F=ma$ and it is known as the equation of Newton's second law of motion.
(b) Mass of the vehicle $=1000\ kg$
Initial velocity $=20\ m/s$
Final velocity $v=0$
Distance travelled $s=50\ m$
(i) On using the third equation of the motion, $v^2=u^2+2as$
$0=20^2+2a\times50$
Or $a=-\frac{400}{100}$
Or $a=-4\ m/s^2$ [-ve sign indicates the retardation]
Therefore, negative acceleraiton or retardation is $4\ m/s^2$
(ii) Force $F=ma$
$=1000\ kg\times 4$
$=4000\ N$
Therefore, $4000\ N$ of the unbalanced force is acted upon the vechicle.
Updated on: 10-Oct-2022
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