Aboy does 465 j of work pulling an empty wagon along level ground with a force of 111 n [31° below the horizontal]. a frictional force of 155 n opposes the motion and is actually slowing the wagon down from an initial high velocity. the distance the wagon travels is "

w = fdcosθ
465 j = (111 n)d(cos31°)
d = (465 j)/(111 n)(cos31°)
d = 4.89 m

where does the frictional force come in here? does it actually affect the distance travelled or does it just slow down the wagon?

Explanation:

When an object is moved over rough floor then during their motion the net work done is

W = 465 J

now the displacement of the wagon can be found by formula of work done as

now here this is the energy which is lost by the boy in order to move the wagon

here frictional force is opposite force while wagon is moving on the rough floor

so here we can say that friction force only oppose the motion due to which object slow down and work done by friction force is loss of energy in form of heat

so here we will say that the work done by boy on the wagon is his energy loss and that energy is used to compensate the energy loss due to friction force.

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)