Chapter 5.1 Review

Me in 1985.

Me today!

1. In which of these cases is work being done? A. a student is holding a boxB. a student lifts a boxC. a jumping deer is in the air going over a fenceD. a pitcher throws the ballE. a volleyball player spikes the ballF. a teacher leans against a wall

1. In which of these cases is work being done? A. a student is holding a boxB. a student lifts a boxC. a jumping deer is in the air going over a fenceD. a pitcher throws the ballE. a volleyball player spikes the ballF. a teacher leans against a wall

2. In each of these cases is the work being done positive or negative? A. a bowler pushes a bowling ballB. gravity slows a ball thrown upward until it stopsC. friction allows a box to slide down a ramp at a constant speed.D. a mother lifts her baby off the floorE. a weightlifter performing the benchpress lowers the weight to his chest F. a catcher applies a force to catch a pitched ball

2. In each of these cases is the work being done positive or negative? A. a bowler pushes a bowling ball - PositiveB. gravity slows a ball thrown upward until it stops - NegativeC. friction allows a box to slide down a ramp at a constant speed. - NegativeD. a mother lifts her baby off the floor -PositiveE. a weightlifter performing the benchpress lowers the weight to his chest - NegativeF. a catcher applies a force to catch a pitched ball - Negative

3. A toddler pulls a wagon with a force of 5 newtons over a displacement of 4 meters. How much work did the toddler do?

Work = force x displacement

W = F x dW = 5 x 4W = 20 J

4. If the wagon in question 3 is pulled over level ground, how much work did gravity do?

ZeroGravity applies no force parallel to the direction of motion. For work to be done a component of the force must be parallel to the direction of motion.

5. If you apply a force of 10 newtons over a displacement of 10 meters, and your friend applies a force of 20 newtons over a displacement of 5 meters, who did the most work?

W = F x dW = 10 x 10 = 100 JW = 20 x 5 = 100 J

The work was the same.

6. If you apply a force of 25 newtons to a box to slide it across a floor a displacement of 8 meters, how much work did you do to the box? If the frictional force on the box was 10 newtons, how much work did friction do to the box over the same 8 meters? What was the net work done on the box?

W = F x dW = 25 x 8W = 200 J

W = 10 x -8W = -80 J

Wnet = 200 + (-80) Wnet = 120 J