Question

A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400m in 20s. Find acceleration. Find the force acting on it if its mass is 7 tonnes. (Hint: 1tonne = 1000 kg)

Answer 1

Hint: The velocity of the body at rest is zero. Use the kinematic equation of motion for calculating the acceleration of the truck. Define what force is, write it in mathematical form and hence find the force acting on the truck. Convert all the units in one system. Use 1tonne=1000kg for converting mass from tonne into kilograms.

Complete step by step solution:
To find the acceleration of the truck:
Let u be the initial velocity of the truck, s be the displacement covered by truck in time t.
Since initially, truck is at rest, we have u=0m/s
From the given data we have, s=400m and t=20s
Using the kinematic equations of motion
\[s=ut+\dfrac{1}{2}a{{t}^{2}}\]
Rearranging the equation to find the acceleration a,
$a=\dfrac{2(s-ut)}{{{t}^{2}}}$
Substituting the given values, we get,
$\begin{align}
  & a=\dfrac{2(400-0\times 20)}{{{20}^{2}}} \\
 & a=2m{{s}^{-2}} \\
\end{align}$
Hence, the acceleration of the truck is $2m{{s}^{-2}}$
To find force:
We know that, in simple words, force is nothing but the product of mass and acceleration. $force=mass\times acceleration$
Mathematically force is given by,
$F=m\times a$
Where,
F= force acting on the body
m= mass of the body
a= acceleration of the body
Conversion of mass into Kg:
$\begin{align}
  & 1tonne=1000Kg \\
 & 7tonne=7\times 1000=7000Kg \\
\end{align}$
We get,
$F=7000Kg\times 1m/{{s}^{2}}=7000N$
Hence, the force acting on the truck is 7000N

Note: Note that the truck rolls down with constant acceleration, therefore you can use the kinematic equation of motion. Be careful about units. We have calculated the mass in kilograms because a standard unit of mass is kilograms (Kg). Do not convert mass in gram until it is required. Convert all the units in one system. Newton, denoted by N is a unit of force. \[1\text{ }newton\text{ }is\text{ }equal\text{ }to=1Kgm/{{s}^{2}}\]