Download Name - East Physical Science

C
Name: ___________________
Period: _____________
1. How is the scientific definition of work different from the everyday definition?
Everyday – something to earn money
Scientific – Force applied to moving an object
2. Describe a situation where a force is applied but no work is done.
Pushing on a wall that doesn’t move. No motion = no work
3. A force of 10,000 N is applied to a stationary wall. How much work is performed?
W=F*d
W = 10,000 N * 0 m
W=0J
4. A 950 N skydiver jumps from an altitude of 3000 m. What is the total work
performed on the skydiver?
W=F*d
W = 950 N * 3000 m
W = 2,850,000 J
5. You are walking from your math class to your science class. You are carrying books
that weigh 20 N (↑). You walk 45 m (→) down the hall, climb 15 m up (↑) the stairs, and
then walk another 30 m (→) to your science class. What is the total work performed on
your books?
* Remember that the movement needs to be in the same direction as the force applied.
W=F*d
W = 20 N * 15 m
W = 300 J
6. How much work is done by a crane that lowers 1000 N of material a distance of 150 m?
W=F*d
W = 1000 N * 150 m
W = 150,000 J
7. A bulldozer performs 80,000 J of work pushing dirt a distance of 16 m. What is the
force of the dirt?
F=W/d
F = 80,000 J / 16 m
F = 5000 N
8. A woman lifts a 35 kg child a distance of 1.5 m (↑) and carries her forward for 6.5 m
(→).
a. How much work does the woman do in lifting the child?
Wt = m * g
Wt = 35 kg * 9.8 m/s/s = 343 N
W=Fxd
W = 343 N x 1.5 m
W = 514.5 J
b. How much work does the child do?
Child applies no force that
causes movement
0J
9. An ant does 1 J of work in dragging a 0.0020 N grain of sugar. How far does the ant
drag the sugar?
d=W/F
d = 1 J / .002 N
d = 500 m
10. How much work is done when a 1 kg mass is raised a vertical distance of 1 m?
F=m*g
F = 1 kg * 9.8 m/s/s = 9.8 N
W=Fxd
W = 9.8 N x 1 m
W = 9.8 J
11. Mrs. Smith pushed a 10 kg desk across a floor for a distance of 5 m (→). She exerted a
horizontal force of 20 N (→). How much work did she do?
W=F*d
W = 20 N * 5 m
W = 100 J
12. A weight lifter lifts a 150 kg barbell above his head from the floor to a height of 2 m.
He holds the barbell there for 5 seconds. How much work does he do during that 5
second interval?
W=F*d
d=0
0J