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# A single force acts on a $3.0\,kg$ particle-like object whose position is given by $x=3.0t-4.0{{t}^{2}}+1.0{{t}^{3}}$, with $x$ in meter and $t$ in seconds. Find the work done by the force from $t=0$ to $t=4.0s$.

Last updated date: 18th Mar 2023
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Hint:We have to use the concept of work energy theorem to solve this question. Work energy theorem is the relation between the work done by the force and corresponding energy.Work done by all the forces on the system is equal to change in kinetic energy of the system.

When we find change in kinetic energy, we need initial and final velocity of the object's.Velocity is rate of change of displacement
$v=\dfrac{dx}{dt}$
In this problem displacement as a function of time
$x=3.0t-4.0{{t}^{2}}+1.0{{t}^{3}}$
Put this displacement in formula of velocity then,
It can be written as
$v=\dfrac{d}{dt}(3.0t-4.0{{t}^{2}}+1.0{{t}^{3}})$

When we differentiate the velocity with respect to time then, we get
$v=3-8t+3{{t}^{2}}$
The object starts moving at $t=0$, it means when we put $t=0$ in the velocity, we get the initial velocity
Put $t=0$in velocity
${{v}_{i}}=3-8\times 0+3\times 0$
$\Rightarrow {{v}_{i}}=3m/s$
Therefore the initial velocity is $3\,m/s$.

The object comes in rest at $t=4$, it means when we put $t=4$in the velocity, we get the final velocity
Put $t=4$in velocity
${{v}_{f}}=3-8\times 4+3\times 4\times 4$
$\Rightarrow {{v}_{f}}=19\,m/s$
Therefore the final velocity is $19\,m/s$. When we calculate the work done, we use work energy theorem
$w=\Delta k$ $........equation1$
Where, $w$ is work done and $\Delta k$ is change in kinetic energy.
S.I Unit of work done is $Joule$.

Now, we calculate change in kinetic energy
$\Delta k=\dfrac{1}{2}m(v_{f}^{2}-v_{i}^{2})$
After putting the value of $m$,${{v}_{i}}$ and ${{v}_{f}}$, we get
$\Delta k=\dfrac{1}{2}\times 3\times [{{(19)}^{2}}-{{(3)}^{2}}]$
$\Rightarrow \Delta k=\dfrac{1}{2}\times 3\times [361-9]$
$\Rightarrow \Delta k=\dfrac{1}{2}\times 3\times 352$
$\Rightarrow \Delta k=528J$
We put the value of $\Delta k$ in equation1
$w=\Delta k$
$\therefore w=528\,J$

Hence, the work done by the force is $528\,J$.

Note:According to law of conservation of energy total mechanical energy is the sum of kinetic energy and potential energy but in this question change in potential energy is zero, so all the mechanical energy is converted into kinetic energy.