Home Page Textbooks Physics Page 216: Lesson Check

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 216: Lesson Check

Exercise 52

Step 1

1 of 2

In this problem, we explain what happens to the power supplied to an object when the rate at which work is done on an object is increased.

Step 2

2 of 2

By definition, the power delivered to an object is the rate at which work is done on the object. Since the rate of work is increased, the power suplied must **increase**.

Exercise 53

Step 1

1 of 4

In this problem, System 1 has force of $F_{1} = 10~mathrm{N}$ and a speed of $v_{1} = 5~mathrm{m/s}$. System 2 has force $F_{2} = 20~mathrm{N}$ and speed $v_{2} = 2~mathrm{m/s}$. We compare their power.

Step 2

2 of 4

The power is the product of the force and speed. For system 1, we have

$$
P_{1} = F_{1}v_{1} = 50~mathrm{W}
$$

Step 3

3 of 4

For system 2, we have

$$
P_{2} = F_{2}v_{2} = 40~mathrm{W}
$$

Step 4

4 of 4

Since we calculated the power, we can see that **system 1** has greater power.

Exercise 54

Step 1

1 of 2

In this power, the power of Engine 1 is twice of the power of Engine 2. We are asked if we can conclude that Engine 1 does twice as much work as Engine 2.

Step 2

2 of 2

The answer is **no**. The work done is the product of the power and time in whic the engine delivers power. We are not given the time of delivering power, so we cannot conclude which one has greater work done.

Exercise 55

Step 1

1 of 2

In this problem, football player 1 exerts twice the force of football player 2. We are asked if we can conclude that plaer 1 produces twice as much power as player 2.

Step 2

2 of 2

The answer is **no**. Another equation for power is $P = Fv$. We are not given the knowledge of the speed, so we cannot conclude which player delivers more power.

Exercise 56

Step 1

1 of 6

In this problem, a bucket of water is raised from the bottom of a well that is $h = 12~mathrm{m}$ deep. The mass of the bucket and water is $m = 5.00~mathrm{kg}$ and takes $t = 15~mathrm{s}$ to teach the top of the well. We calculate the power required. We use $g = 9.81~mathrm{m/s^{2}}$.

Step 2

2 of 6

We assume that the force lifting the bucket is at least the weight of the bucket and water. We have

$$
F = mg
$$

Step 3

3 of 6

The net force is zero, so the speed must be constant. The speed must be

$$
v = frac{h}{t}
$$

Step 4

4 of 6

The power must be

$$
begin{aligned}
P &= Fv \
&= mg frac{h}{t} \
&= left( 5.00~mathrm{kg} right) left( 9.81~mathrm{m/s^{2}} right) frac{12~mathrm{m}}{15~mathrm{s}} \
&= 39.24000~mathrm{W} \
P &= boxed{39~mathrm{W} }
end{aligned}
$$

Step 5

5 of 6

The power must be

Result

6 of 6

$$
P = 39~mathrm{W}
$$

Exercise 57

Step 1

1 of 3

In this problem, a fly of mass $m = 1.4 times 10^{-3}~mathrm{kg}$ alks straight up a windowpane at rate $v = 2.3 times 10^{-3}~mathrm{m/s}$. We calculate its power output. We use $g = 9.81~mathrm{m/s^{2}}$.

Step 2

2 of 3

The fly exerts a force equal to its weight in order to move in constant speed. The power output must be

$$
begin{aligned}
P &= Fv \
&= mgv \
&= left( 1.4 times 10^{-3}~mathrm{kg} right) left( 9.81~mathrm{m/s^{2}} right) left(2.3 times 10^{-3}~mathrm{m/s}right) \
&= 3.15882 times 10^{-4}~mathrm{W} \
P &= boxed{3.2 times 10^{-4}~mathrm{W}}
end{aligned}
$$

Result

3 of 3

$$
P = 3.2 times 10^{-4}~mathrm{W}
$$

Exercise 58

Step 1

1 of 3

In this problem, a kayaker paddles with a power output of $P = 50.0~mathrm{W}$ to maintain a steady speed of $v = 1.50~mathrm{m/s}$. We calculate the force exerted by the kayaker.

Step 2

2 of 3

From the equation of power, we have

$$
begin{aligned}
P &= Fv \
implies F &= frac{P}{v} \
&= frac{50.0~mathrm{W}}{1.50~mathrm{m/s}} \
&= 33.33333~mathrm{N} \
F &= boxed{33.3~mathrm{N}}
end{aligned}
$$

Result

3 of 3

$$
F = 33.3~mathrm{N}
$$

Exercise 59

Step 1

1 of 3

In this problem, an ice cube is palced in a microwave oven. The oven delivers a power of $P = 105~mathrm{W}$ and the ice cube takes $W = 32200~mathrm{J}$ to melt. We calculate the time it takes for the ice cube to melt.

Step 2

2 of 3

From the definition of power, we can calculate $t$.

$$
begin{aligned}
P &= frac{W}{t} \
implies t &= frac{W}{P} \
&= frac{32200~mathrm{J}}{150~mathrm{W}} \
&= 306.66667 \
t &= boxed{307~mathrm{s}}
end{aligned}
$$

Result

3 of 3

$$
t = 307~mathrm{s}
$$

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