## 1) DeAnna, with a mass of 60.0 kg, climbs 3.5 m up a gymnasium rope. How much energy does a system containing DeAnna and Earth gain fr

Question

1) DeAnna, with a mass of 60.0 kg, climbs 3.5 m up a gymnasium rope. How much energy does
a system containing DeAnna and Earth gain from this climb?
2) An electric motor develops 65 kW of power as it lifts a loaded elevator 17.5 m in 35 s. How
much force does the motor exert?
3) A 25.0-kg shell is shot from a cannon at Earth’s surface. The reference level is Earth’s surface.
a. What is the shell-Earth system’s gravitational potential energy when the shell’s height is
425 m?
b. What is the change in the system’s potential energy when the shell falls to a height of 225
m?
4) Shawn and his bike have a combined mass of 45.0 kg. Shawn rides his bike 1.80 km in 10.0
min at a constant velocity. What is the system’s kinetic energy?
5) Mary weighs 505 N. She walks down a 5.50-m-high flight of stairs. What is the change in the
potential energy of the Mary-Earth system?
6) Katia and Angela each have a mass of 45 kg and are moving together with a speed of 10.0
m/s.
a. What is their combined kinetic energy.
b. What is the ratio of their combined mass to Katia’s mass?
c. What is the ratio of their combined kinetic energy to Katia’s kinetic energy? Explain how
this ratio relates to the ratio of their masses.
7) A 10.0-kg test rocket is fired vertically. When the engine stops firing, the rocket’s kinetic
energy is 1960 J. After the fuel is burned, to what additional height will the rocket rise?

in progress 0
3 years 2021-08-09T05:21:32+00:00 1 Answers 410 views 0

1. 1. As he climbs… He’s gaining potential Energy.

PE=mgh = 60×9.8×3.5=2058J. This is the potential Energy gain.

2. Power = Workdone/Time.

65kW=65,000watts

65000=Workdone/35

Workdone=65,000 x 35=2,275,000J

Recall… Workdone=Force x Distance

given distance =17.5m

Force =Workdone/distance

=2,275,000/17.5

Force=130,000N or 130kN.

3.Potential Energy Increases or decreases as height increases or decreases respectively.

a) PE=mgh =25×9.8×425=104,125J.

b) change in PE=mg(h°-h) =25 x 9.8(225-425)

= -49000J. The Energy is negative because the system loses 49000J.

4. At Constant velocity; d=vt

where d=distance,v=velocity, t=time

v=d/t

1.8km=1800m

10min =10×60=600seconds

v=1800/600

v=3ms-¹

KE=1/2Mv²

=1/2 x 45 x 3² =202.5J.

5.Mary weighs 505N.

Weight=Mass x acceleration due to gravity

W=mg

505=m x 9.8

m=505/9.8 =51.53kg.

Since Mary is coming down… she’s decreasing her potential energy cos Potential Energy Decreases as you decrease your height. So Her Initial height was 5.5m… she alighted to the ground which we’ll say is the ground(h=0)

Change in PE=mg(h°-h)=51.53 x 9.8(0-5.5) = -2777.5J.

This energy is negative because she lost potential Energy.

6. Since each has a Mass of 45kg… and they’re moving together…. Their combined Mass =45+45=90kg.

Velocity given as =10ms-¹

a)Combined KE=1/2MV² = 1/2 x 90 x 10² =4500J.

b)Ratio of combined Mass to Katias mass = CM/KM

Where CM=Combined Mass

KM=Katia’s mass

=90/45 =2:1

c) Katia’s kinetic E =1/2 x 45 x 10² =2250J.

Ratio of combined to Katia’s = CK/KK

CK=Combiner Kinetic Energy

KK=Katia’s kinetic energy

= 4500/2250 =2:1.

This answer should be the same cause from the Formula KE=1/2mv²

KInetic Energy varies Directly with Mass

so If the ratio of Masses is 2:1… That of KE should be the same because of their direct Relationship✌.

7.The height at which the Rocket Engine stops is not how far it went. Theres a Minute height that it’ll still go before it starts going back down.

from the question… The energy when the engine stops is 1960J

PE=mgh

1960=10×9.8xh

h=1960/98 =20m.

The engine stops at this height

but The additional Minute height that was added before itll start falling cant be calculated from the insufficient info given in this question. If we had the Energy at the additional height-added… We can calculate that height and then subtract 20 from it. That will give the additional added height.

Thats pretty Much Everything✌

Spectro… Out