Question

The human small intestine moves food along its 6.9- m length by means of peristalsis, a contraction and relaxation of muscles that propagate in a wave along the digestive tract. Part A If the average force exerted by peristalsis is 0.18 N , how much work does the small intestine do to move food along its entire length?

Answer 1

Answer:

1.24 J

Explanation:

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the directions of F and d

In this problem,

F = 0.18 N is the average force exerted by the muscles

d = 6.9 m is the displacement

$\theta=0^{\circ}$ (we assume the direction of the force is parallel to the displacement)

Substituting,

$W=(0.18)(6.9)(cos 0)=1.24 J$