Question

An object of mass $2kg$ is dropped from a tower which hits the ground in $10s$. Calculate the work done

Answer 1

Hint:here in the question they have given the mass $m$that is $2kg$and they have also mentioned the time $t$ that is $10s$by using these two data we have to find the work done
As we know the work done can be defined as the transferring of energy from an object via the application of force along a displacement.

Complete step by step solution:
As we know the formulae for calculating the work done that is

$W = Fs........(1)$

Here $F$represent the force and $s$represent the displacement

But in the question we have mass and time so first we have to find the value of force and displacement

So to find the value of force we have formulae

$F = ma$

Here $m$is the mass and \[a\]represent the acceleration due to gravity
Hence we know the value of $m$that is $2kg$and the acceleration due to gravity is $9.8m/{s^2}$that is constant

$F = 2kg \times 9.8m/{s^2}$

After simplifying we get

$F = 19.6N$

Now we have to calculate the value of displacement $s$

So we know the formulae to calculate the displacement that is

\[s = \dfrac{1}{2}a{t^2}\]

Here \[a\]represent the acceleration due to gravity that is $9.8m/{s^2}$ and $t$ is the time that is $10s$

Now substitute the values in the above equation

$s = \dfrac{1}{2} \times 9.8m/{s^2} \times 10s$

After simplifying the equation we get

$s = 49m/s$

Now substitute the value of displacement $s$and the force $F$in equation $(1)$

$W = 19.6 \times 49$

After calculating we get the value of work done

$W = 960.4N - m$

Note:a force can be defined as the motion of an object. A force can cause an object with mass to change its velocity, force has both magnitude and direction
 Displacement is a vector quantity whose length is the shortest distance from the initial to the final position .