A particle moves a distance of $190 \mathrm { ft }$ along a straight line. As it moves, it is acted upon by a constant force of magnitude $15 \mathrm { lb }$ in a direction opposite to that of the motion. What is the work done by the force?
A) $- \frac { 3 } { 38 } \mathrm { ft } - \mathrm { lb }$
B) $- 2,850 \mathrm { ft } - \mathrm { lb }$
C) $- \frac { 1 } { 2850 } \mathrm { ft } - \mathrm { lb }$
D) $- \frac { 38 } { 3 } \mathrm { ft } - \mathrm { lb }$

Question

Quizplus · Accepted Answer

The work done by a force is given by the formula $W = Fd\cos	heta$, where $F$ is the magnitude of the force, $d$ is the distance moved, and $	heta$ is the angle between the force and the direction of motion. In this case, the force and the direction of motion are in opposite directions, so $	heta = 180^\circ$. Thus, we have $W = (15\mathrm{lb})(190\mathrm{ft})(\cos 180^\circ) = -2,850\mathrm{ft}-\mathrm{lb}$. Note that the negative sign indicates that the force is doing negative work, i.e. removing energy from the particle.

A Particle Moves a Distance Of 190ft190 \mathrm { ft }190ft

A Particle Moves a Distance Of $190 \mathrm { ft }$