Question

A truck starts from rest and rolls down the hill with constant acceleration. It travels a distance of 500 m in 25 seconds. Find the force acting on it if its mass is 6 metric tons.

Answer 1

We know that the force acting on a particle is equal to the rate of change of momentum of a particle. First, we need to find the acceleration. Since in this problem given initial velocity, time and distance, we can use Newton's equation of motion, $s = ut + \dfrac{1}{2}a{t^2}$ to find the acceleration. Then using Newton's second law we find the force acting on it for a given mass.

Complete step by step answer:
For calculation of the force acting on it when mass is given. We use the Newton’s equation of motion i.e.,
$s = ut + \dfrac{1}{2}a{t^2}$---(1)
where $u$ is the initial velocity, $s$ is the distance, $t$ is the time taken to cover a distance and $a$ is the acceleration.
Here, it is given that a truck starts from rest and covers a distance of 500 m in 25 seconds.
Hence given $u = 0$,$s = 500\;m$ and $t = 25\;\operatorname{s}$.
Them the equation (1) becomes
$500 = 0 + \dfrac{1}{2}a{\left( {25} \right)^2}$
$a = \dfrac{{500 \times 2}}{{25 \times 25}} = \dfrac{{40}}{{25}} = 1.6$
Hence acceleration a = 1.6$$$$m/{s^2}-----(2)

For calculation the force acting on it when mass is given. We use the newton’s second law of motion i.e.,
$F = ma$--(3)
Where, $F$ is force and $m$ is the mass on which force is acting.
Given $m = 6$ $metric\;tons$. Since, we know that $1\;metric\;tons = 1000\;kg$. Then $m = 6\;metric\;tons = 6000\;kg$.
Then using the value of $a$from the equation (2) in the equation (3), we get
$\therefore F = 6000 \times 1.6 = 9600$ $N$

Hence the force acting on a truck if its mass is 6 metric tons is 9600\;kg\,m\,{s^{ - 2}} = $$$$9600$$$$N.

Note: We must be taken care of using the units. For this problem mass is given in metric tons but mass is always taken in kg. So, we must convert metric tons into kg. Also note that for any question of this type, we use Newton's three equations of motion. If $u$ is the initial velocity and $v$ is the final velocity of the object, $s$ is the distance travelled by the object in $t$ seconds with acceleration $a$ then the three equations of motion are $s = ut + \dfrac{1}{2}a{t^2}$, $v = u + at$, ${v^2} - {u^2} = 2as$ . which equation is used will depend upon the given conditions of the questions.