*Whee!*(Video link: "Human Slingshot Slip and Slide - Vooray.")

*KE*is the energy of motion. (We'll deal with

_{tr}*rotational*kinetic energy

*KE*later on.)

_{rot}*KE*depends on the mass

_{tr}*m*and the square of the speed

*v*of the object, and the resulting units of kg·m

^{2}/s

^{2}are also expressed as joules. A stationary object has no translational kinetic energy, and the faster an object moves, the more translational kinetic energy it has.

Instead of finding out how much translational kinetic energy

*KE*an object has, often we are more concerned with its initial-to-final change ∆

_{tr}*KE*, which is the final amount of translational kinetic energy minus the initial amount of translational kinetic energy. Notice how the common factors of (1/2) and mass

_{tr}*m*(which is presumed to be constant) are pulled out, leaving a "difference of squares" for the final and initial speeds in the parenthesis.

*W*either on, or against the motion of the object (here the loaded sledge the horses are pulling).

*s*(note the unusual use of "

*s*" for a generic ∆

*x*, ∆

*y*, or other direction displacement!), and the angle between the force and displacement vectors (when placed tail-to-tail) is anything besides 90°. The units of work are given in N·m, or yet again, joules, so keep in mind that work can be done on or against any mechanical energy form or forms.

If the force does work

*on*the object (by being exerted along the direction of its motion), then the work will have a

*positive*sign, and the translational kinetic energy of the object will

*increase*, making the sign of the ∆

*KE*term

_{tr}*positive*. Note for this case how the left- and right-hand sides of this equation must have the same

*positive*sign.

If the force does work

*against*the object (by being exerted opposite to the direction of its motion), then the work will have a

*negative*sign, and the translational kinetic energy of the object will

*decrease*, making the sign of the ∆

*KE*term

_{tr}*negative*. Note for this case how the left- and right-hand sides of this equation must have the same

*negative*sign.

*Squirrel catapult!*(Note the person behind the sliding glass door, cutting the release cord with a scissors to launch the squirrel.) (Video link: "squirrelcatapult.gif.")

To keep things simple, let's consider a strictly horizontal version of this contraption, which would make the work done by any vertical forces (such as the weight force of Earth on the squirrel) zero, as the angle between these vertical forces and the horizontal direction of motion is 90°.

The bungee cord (and basket) exerts a force on the squirrel directed to the right, along the direction of motion, so the bungee cord does work

*on*the squirrel. As a result, the squirrel picks up speed (starting from rest), and since translational kinetic energy depends on the square of the speed, since speed increases, then the squirrel's translational kinetic energy increases.)

In the work-energy theorem equation:

*W*= ∆

_{}*KE*,

_{tr}the work will have a

*positive*sign (as work was done on the squirrel by the bungee cords), causing the squirrel's translational kinetic energy to increase (and also have a

*positive*sign), so the +/– signs of both the left-hand side and the right-hand side of the equation are consistent with each other:

(+) = (+).

(If you were to calculate the numerical values for the work done (in J) and the resulting numerical value for the increase in translational kinetic energy (in J), then they would have to be equal to each other (along with having the same sign).

For the catapulted squirrel, the bungee cord force does work __________ the squirrel, which __________ the squirrel's translational kinetic energy.)

(A) on; increases.

(B) against; decreases.

(C) (Unsure/lost/guessing/help!)

For the braking car, the brakes do work __________ the car, which __________ the car's translational kinetic energy.

(A) on; increases.

(B) against; decreases.

(C) (Unsure/lost/guessing/help!)

For Mrs. P-dog being catapulted upwards, the bungee cords do work __________ Mrs. P-dog, while the weight force does work __________ Mrs. P-dog.

(A) on; on.

(B) on; against.

(C) against; on.

(D) against; against.

(E) (Unsure/lost/guessing/help!)

For Mrs. P-dog's translational kinetic energy to be increased while being catapulted upwards, the amount of work from the bungee cords must be __________ the amount of work from the weight force.

(A) less than.

(B) the same as.

(C) greater than.

(D) (Not enough information is given.)

(E) (Unsure/lost/guessing/help!)