The work done by an applied variable force $F=x +x^{3}$ from $x=o m$ to $x=2, m$ where $x$ is displacement, is
- 6J
- 8J
- 10J
- 12J
The Correct Option is A
Solution and Explanation
and $x =0 m$ to $x =2 m$
We know
$W=\int F d x$
$W =\int\limits_{0}^{2}\left(x +x^{3}\right) d x$
$W=\left[\frac{x^{2}}{x^{4}} 4\right]_{0}^{2}$
$W=\left[\frac{4}{2}+\frac{16}{4}\right]$
$W=6\, J$
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Concepts Used:
Work
Work is the product of the component of the force in the direction of the displacement and the magnitude of this displacement.
Work Formula:
W = Force × Distance
Where,
Work (W) is equal to the force (f) time the distance.
Work Equations:
W = F d Cos θ
Where,
W = Amount of work, F = Vector of force, D = Magnitude of displacement, and θ = Angle between the vector of force and vector of displacement.
Unit of Work:
The SI unit for the work is the joule (J), and it is defined as the work done by a force of 1 Newton in moving an object for a distance of one unit meter in the direction of the force.
Work formula is used to measure the amount of work done, force, or displacement in any maths or real-life problem. It is written as in Newton meter or Nm.
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