Question

A force in the positive direction of an x axis acts on a 4.83kg 4.83 kg block moving along that axis. If the magnitude of the force is =10.3−x/2.6 N F = 10.3 e - x / 2.6 ⁢ N , with x x in meters, find the maximum amount of work this force can do on the block as it begins its motion at x=0.00 x = 0.00 , moving in the positive direction of the x axis. Wmx= W m a x = Type your answer here J J .

Answer 1

The maximum work done by the force on the block as it begins motion at x=0.00 and moves in the positive direction is approximately 26.78 joules, assuming no other forces like friction are acting on the block and the force decays exponentially.

Step-by-step explanation:

To find the maximum work done by a force given by F = 10.3e-x/2.6, we need to integrate the force over the distance it acts on the 4.83 kg block. The work done, W, is the integral of F to x, from the initial position to where the force effectively becomes zero or close to it (as it decreases exponentially). However, the maximum work on a block moving in the positive direction of the x-axis would involve taking the limit of x to infinity since the force decreases but never truly reaches zero. Therefore:

W = ∫ F dx

Before integrating, we apply the limit as x approaches infinity to determine the maximum possible work. For the force function F = 10.3e-x/2.6, we have:

Wmax = ∫0∞ 10.3e-x/2.6 dx

Using the properties of exponential functions and integrals, the work simplifies to:

Wmax = (10.3)(2.6), which simplifies to approximately 26.78 joules (J).

This calculation assumes no other forces are acting on the block such as friction, and it relies on the fact that a decaying exponential function approaches zero but integrates to a finite value over an infinite interval.