Question 8 A 100 N for is applied to move a crate 25 m which encounters 37 N of friction. What is the negative amount of work done in joules? square 1 pts

The work done by friction is given by the formula:<br /><br />Work = Force x Distance x cos(theta)<br /><br />Where:<br /><br />* Force is the frictional force (37 N)<br />* Distance is the distance the crate moves (25 m)<br />* Theta is the angle between the force and the direction of motion. Since friction always opposes motion, the angle is 180 degrees.<br /><br />cos(180°) = -1<br /><br />Therefore, the work done by friction is:<br /><br />Work = (37 N) * (25 m) * (-1) = -925 Joules<br /><br />So the negative amount of work done is $\boxed{-925}$ Joules.<br />