**Problem 5A**

**Work ANSWERS!!**

**Fill each blank below with the word or phrase that completes
the statement.**

1. The total work done on
an object is the _** sum**__of
the work done

by the individual forces.

2. Positive work is done
when____** force is in direction of motion**____

.

3. The work done by a force
parallel to the displacement is determined by

multiplying the _** amount**___of the force by the__

4. The difference between
positive and negative work is __** pos work is in the direction of motion,
negative work is opposite the direction of motion**_______

.

5. Positive and work both
occur parallel to the __** displacement (motion)**____.

6. No work is done when the
_** force**_____is perpendicular
to the___

7. What is the equation for
determining the amount of work done on an

object when the force is
applied at an angle? __** Work = Force *cos (θ) *Dis or F*D*cos(θ)**_

**12. A wagon is pulled 45
m along a level road at constant velocity. Find the

amount of work done on the
wagon by a force of 85 N that is applied to

the handle and that makes
an angle of 20.0° with the horizontal.

*Work = Force *cos
(θ) *Dis*

*Work = 85 N* cos (20)
* 45*

* =
79.87 N* 45 m = 3594 N m or 3594 Joules*

13. A piano is lifted 3.0 m
vertically. Determine the work done on the piano

if its mass is 750 kg.

* Work = F *D =Fg *D = mg h = 750 (9.8)
(3) = 7350 N * 3 m= 22050 Joules*

**14. Determine the work
done on a sled that is pulled 20.0 m by a 105 N

force applied at an angle
of 50° to the horizontal.

*Work = Force *cos
(θ) *Dis*

* =
105 * cos (50) * 20 = 67.5 N* 20 m =1349.9 Joules*

15. A 34.5 kg box with an
initial velocity of 10.0 m/s slides to a stop along a

level road. If the
displacement of the box is 17.5 m, determine the force

of friction on the box and
the work done to stop it.

* Work = Force * Dis = mA *Dist*

*We need A……*

* Vf ^{2}=Vi^{2}
+ 2 A D A=Vi^{2}/(2D)=
10(10)/(2*17.5) =100/35=-2.857 m/s^{2}=A*

*Ff=mA= 34.5 kg (2.857
m/s ^{s}) = - 98.57 N = Ff*

*Work = F*D= -98.57
N*17.5m= 1725 Joules*

WORK

The largest palace in the
world is the Imperial Palace in Beijing, China.

The palace covers a
rectangle 750 m long by 960 m wide. If you were to

push a lawn mower around
the perimeter of such an area, applying a constant

horizontal force of 60.0
N,what amount of work would you do?

*Perimeter is 750
+960+750+960= 3420 m*

*Work = F*D=
60*3420=205, 200 Joules*

1. With an overall height
of 195 m, Lake Point Tower in Chicago is the

tallest apartment building
in the United States (although not the tallest

building in which there are
apartments). Suppose you live on the top

floor of the building and
your mass is 60.0 kg. How much work is done

on you by the force of
gravity as you ride the elevator from the top

floor to the ground floor?

*Work=F*D = Fg*D= mg h
= 60(9.8) * 195= 588 N * 195 m = - 114, 600 Joules=Work done by gravity*

2. In 1985 in San Antonio,
Texas, an entire hotel building was moved several

blocks on 36 dollies. The
mass of the building was about

1.45 *10^{6} kg.
Suppose the amount of work done on the building was

100 Mega J and the
resistive force that had to be overcome was just

2.00 percent of the
building’s weight. How far was the building moved?

*Mass = 1 450 000 kg,
Weight =mg= 1 450 000 * 9.8 = 14 210 000 Newtons*

*Resistive Force = 2%
(Weight) = .02 ( 14 210 000 ) =284 200 Newtons*

*Work = F * D= 100,
000, 000 Joules = 284 200 N* Dist*

*Dist = Work/Force =
100 000 000 / 282 200 = 351.9 m*

* *

3. A hummingbird has a mass
of about 1.7 g. If the hummingbird ascends

straight up with a net
acceleration of 1.2 m/s^{2}, how much work

does it do over a distance
of 8.0 m?

*Work = Fup * Dist*

* *

**Fnet = Fup – Weight,
**

**Fup = Weight + Fnet =
.0017*9.8 + .0017*1.2**

**Fnet = mA= .0017*1.2
=.00204 Newtons**

**Weight= mg = .0017*9.8 =
.01666 Newtons**

**Fup = Weight + Fnet =
.00104+.01666= .0187 Newtons**

**Work = Fup* Height= .0187
N * 8 m = .1496 Joules=Work**