The work done by a force F = ($-6x^3 $)i N, in displacing a particle from x = 4 m to x = - 2 m is
- 360
- 240 J
- -240 J
- -360 J
The Correct Option is D
Solution and Explanation
Given
$F = -(6x^3 )iN$
$W=-\int ^4_{-2} 6x^3 dx$
$W=-6\left[\frac{x^4}{4}\right]^4_{-2}$
$W=-\frac{6}{4}[(4)^4-(-2)^4]$
$W=-\frac{6}{4}[256-16]$
W=-1.5 $\times$ 240 =-360 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.