Home Page Textbooks Physics: Principles and Problems Page 265: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

Textbook solutions

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Page 265: Section Review

Exercise 15

Step 1

1 of 4

Murimi pushes a mass across a floor. It is given that mass id 20 kg, distance is 10 m and horizontal force is 80 N. We have to calculate amount of work done by Murimi.

Step 2

2 of 4

Work is equal to force times distance:

$$
W=Fs
$$

Step 3

3 of 4

When we put numbers in we get:

$$
W=80cdot10
$$

$$
boxed{W=800,,rm J}
$$

Result

4 of 4

$$
W=800,,rm J
$$

Exercise 16

Step 1

1 of 4

A mover loads a refrigerator into a moving van by pushing it up. It is given that mass of al refrigerator is 185 kg, height is 10 m and angle of inclination is 11$^o$ and there is no friction. We have to calculate how much work is done by the mover.\

Step 2

2 of 4

To calculate work we can use equation:
$$W=Fs$$
in this case force is equal to weight of the refrigerator and distance is equal to height.
$$W=mghcdot sin (theta )$$

Step 3

3 of 4

When we put numbers in we get:
$$W=185cdot9.81cdot10 cdot sin (11,,^{o} )$$
$$boxed{W=3460,,rm J}$$

Result

4 of 4

$$W=3460,,rm J$$

Exercise 17

Step 1

1 of 1

Work is equal to force times distance:

$$
W=Fs
$$

We can see that work only depends on force and distance and if we change how fast we move the book it will not change the work done on the book, answer on this question is $textbf{no}$ it does not depend on how fast you rise it.

Power is defined as work per time:

$$
P=frac{W}{t}
$$

Now we can see that power depends on time and if we move the book faster time will decrease, answer on this question is $textbf{yes}$ it does depend on how fast you rise it.

Exercise 18

Step 1

1 of 4

We have to calculate how much power does the elevator generates when it lifts a total mass of $1.1cdot10^{3},,rm kg$ a distance of 40 m in 12.5 s.

Step 2

2 of 4

First we have to calculate total work for that, we can do that with equation:
$$W=Fs$$
in this case force is equal to weight and distance is equal to height.
And now when we have work we can get power:
$$P=frac{W}{t}$$
When we connect those equations we get:
$$P=frac{mgh}{t}$$

Step 3

3 of 4

When we put numbers in equation we get:
$$P=frac{1.1cdot10^{3}cdot9.81cdot 40}{12.5}$$
$$boxed{P=34530,,rm W}$$

Result

4 of 4

$$P=34530,,rm W$$

Exercise 19

Step 1

1 of 4

We have to calculate how much work does the force of gravity do on the ball when a 0.18 kg ball falls 2.5 m.

Step 2

2 of 4

We can calculate work with equation:

$$
W=Fs
$$

in this case force is equal to weight and distance is equal to height.

$$
W=mgh
$$

Step 3

3 of 4

When we put numbers in we get:

$$
W=0.18cdot9.81cdot2.5
$$

$$
boxed{W=4.41,,rm J}
$$

Result

4 of 4

W=4.41,,rm J

Exercise 20

Step 1

1 of 4

A forklift raises a box 1.2 m and does 7 kJ of work on it. We have to calculate the mass of the box.

Step 2

2 of 4

We can use equation for work to calculate this:

$$
W=Fs
$$

In this case force is equal to weight and distance is equal to height.

$$
W=mghrightarrow m=frac{W}{gh}
$$

Step 3

3 of 4

When we put numbers in we get:

$$
m=frac{7cdot10^{3}}{9.81cdot1.2}
$$

$$
boxed{m=595,,rm kg}
$$

Result

4 of 4

m=595,,rm kg

Exercise 21

Step 1

1 of 2

You and a friend each carry identical boxes
from the first floor of a building to a room located
on the second floor, farther down the hall. You
choose to carry the box first up the stairs, and then
down the hall to the room. Your friend carries it
down the hall on the first floor, then up a different
stairwell to the second floor. Question is who does more work.

Step 2

2 of 2

To answer this question we can use definition of work:

$$
W=vec Fcdotvec s
$$

In this case to lift a box we have to overcome gravitational force. That means if we go down our force on the box will be in opposite direction that we are moving and work will be negative, if we go up our force will be in the same direction and work will be positive, this means that it does not matter if we move in any direction as long we and up in the same spot total work will be the same. Answer to this question is you and your friend will do the same work on the box.

Exercise 22

Step 1

1 of 2

Question is if the work done on an object doubles its kinetic energy, does it double its velocity.

We can write relation between velocity and kinetic energy with equation:

$$
E_k=frac{mv^2}{2}rightarrow v=sqrt{frac{2E_k}{m}}
$$

now we can see that if kinetic energy is doubled, velocity is not. When we put that $E_{k2}=2E_k$ we get:

$$
v_2=sqrt{frac{2E_{k2}}{m}}=sqrt{frac{2cdot2E_{k}}{m}}=sqrt{2}sqrt{frac{2E_{k}}{m}}
$$

$$
boxed{v_2=sqrt{2}v}
$$

Result

2 of 2

$$
v_2=sqrt{2}v
$$

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Page 265: Section Review

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