A variable force acts on a 2.0 -kilogram mass, displacing...

A variable force acts on a 2.0 -kilogram mass, displacing the mass 5.0 meters. The force and displacement are graphed above. Determine the work done on the mass by the variable force.
(A) 5 $\mathrm{J}$
(B) 10 $\mathrm{J}$
(C) 15 $\mathrm{J}$
(D) 20 $\mathrm{J}$
(E) 30 $\mathrm{J}$

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

Work is defined as the energy transferred by a force acting over a distance. For any force acting on an object, work is calculated as the dot product of the force and the displacement. When the force and the displacement are in the same direction, the work is simply the product of the magnitude of the force and the displacement.

Variable Force

When a force varies with position, the work done is calculated by summing up the incremental contributions over the path of motion. This is done through integration, which computes the total effect of a variable force over a displacement.

Integration in Work Calculations

In the context of a variable force, the work done is determined by integrating the force with respect to displacement over the interval of motion. This integral effectively accumulates all the small amounts of work done throughout the displacement.

Graphical Interpretation (Area Under the Curve)

When the force versus displacement is graphed, the work done by the force is represented by the area under the curve between the limits of the displacement. Evaluating this area provides the same result as integrating the force function over the distance traveled.

Transcript

Please give Ace some feedback