A force F=(7-2 x+3 x^2) N applied on a 2 kg mass which displaces it from x=0 to x=5 m. Work do

step 1 In this question we have given force that is applied on a 2kG mass and we have given the displac

A force F=(7-2 x+3 x^2) N applied on a 2 kg mass which...

A force $\mathrm{F}=\left(7-2 x+3 x^{2}\right) \mathrm{N}$ applied on a $2 \mathrm{~kg}$ mass which displaces it from $x=0$ to $x=5 \mathrm{~m}$. Work done in joule is:
(a) $70 \mathrm{~J}$
(b) $270 \mathrm{~J}$
(c) $35 \mathrm{~J}$
(d) $135 \mathrm{~J}$

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Work Done by a Force

In physics, work is the measure of energy transfer when a force acts upon an object to cause displacement. When dealing with variable forces, work is determined by summing the contributions over each infinitesimal displacement, which is typically carried out using integration.

Variable Force and Force as a Function of Position

A variable force depends on the position of the object and is often represented as a function of that position (F(x)). This concept is crucial because it acknowledges that the force can change in magnitude (and sometimes direction) along the path of motion, requiring different methods than those used for constant forces.

Definite Integration in Physics

Definite integration is a calculus technique used to compute the total accumulation of a quantity when its rate of change is variable. In the context of work done by a force, it involves integrating the force function with respect to position over a specific interval to find the total work performed on an object.

Transcript

Please give Ace some feedback