Question 1

Solve the problem.
-A fisherman is about to reel in a 9-lb fish located 12 ft directly below him. If the fishing line weighs 1 oz per foot, how much work will it take to reel in the fish? Round your answer to the nearest tenth, if necessary.
A) 120 ft · lb
B) 180 ft · lb
C) 117 ft · lb
D) 112.5 ft · lb

Accepted Answer

The answer of Solve the problem.
-A fisherman is about to...

Question 2

Solve the problem.
-A construction crane lifts a 100-lb bucket originally containing 130 lb of sand at a constant rate. The sand leaks out at a constant rate so that there is only 65 lb of sand left when the crane reaches a height of 60 feet. How
Much work is done by the crane?
A) 13,800 ft · lb
B) 5850 ft · lb
C) 2550 ft · lb
D) 11,850 ft · lb

Accepted Answer

The answer of Solve the problem.
-A construction crane lifts a...

Question 3

Solve the problem.
-A bathroom scale is compressed $\frac { 1 } { 4 }$ in. when a $220 \mathrm { lb }$ person stands on it. Assuming that the scale behaves like a spring that obeys Hooke's law, how much does someone who compresses the scale $\frac { 1 } { 8 }$ in. weigh?
A) 110 lb
B) 55 lb
C) 440 lb
D) 165 lb

Accepted Answer

Hooke's law states that the force $ F $ needed to extend or compress a spring by some distance $ x $ is proportional to that distance. That is, $ F = kx $, where $ k $ is a constant specific to the spring. In this case, when the scale is compressed $\frac{1}{4}$ inch, it supports a weight of 220 lb. This means $ 220 = k \times \frac{1}{4} $. To find the weight of a person who compresses the scale $\frac{1}{8}$ inch, we first find $ k $ and then use it to find the new weight. Since $ k = \frac{220}{\frac{1}{4}} = 880 $, the weight for a $\frac{1}{8}$ inch compression is $ F = 880 \times \frac{1}{8} = 110 $ lb.

Question 4

Solve the problem.
-A vertical right circular cylindrical tank measures $20 \mathrm { ft }$ high and $12 \mathrm { ft }$ in diameter. It is full of oil weighing $60 \mathrm { lb } / \mathrm { ft } ^ { 3 }$. How much work does it take to pump the oil to the level of the top of the tank? Give your answer to the nearest $\mathrm { ft } \cdot \mathrm { lb }$.
A) 22,619 ft · lb
B) 5,428,672 ft · lb
C) 2,714,336 ft · lb
D) 1,357,168 ft · lb

Accepted Answer

The answer of Solve the problem.
-A vertical right circular cylindrical...

Question 5

Solve the problem.
-It takes a force of 12,000 lb to compress a spring from its free height of 15 in. to its fully compressed height of 10 in. How much work does it take to compress the spring the first inch?
A) 600 in.-lb
B) 1200 in.-lb
C) 2400 in.-lb
D) 120,000 in.-lb

Accepted Answer

The work done to compress a spring is given by the formula:
W = (1/2)kx^2
where W is work done, k is the spring constant, and x is the displacement.

To find the spring constant, we can use the given information:
12,000 lb = k(15 in. - 10 in.)
k = 2,400 lb/in.

Now, we can use this value of k to find the work done to compress the spring by 1 in:
W = (1/2)(2,400 lb/in.)(1 in.)^2
W = 1,200 in.-lb

Question 6

Solve the problem.
-The base of a rectangular tank measures $7 \mathrm { ft }$ by $14 \mathrm { ft }$. The tank is $17 \mathrm { ft }$ tall, and its top is $10 \mathrm { ft }$ below ground level. The tank is full of water weighing $62.4 \mathrm { lb } / \mathrm { ft } ^ { 3 }$. How much work does it take to empty the tank by pumping the water to ground level? Give your answer to the nearest $\mathrm { ft }$ ' $\mathrm { lb }$.
A) 883,646 ft · lb
B) 3,690,523 ft · lb
C) 883,816 ft · lb
D) 1,923,230 ft · lb

Accepted Answer

The work required is calculated by integrating the weight of each water slice times the distance it needs to be lifted. The total work is $ \int_0^{17} 62.4 \times 98 \times (27-y) \, dy = 1,923,230 \, \text{ft} \cdot \text{lb} $.

Question 7

Solve the problem.
-Find the work done in winding up a 175-ft cable that weighs 3.00 lb/ft.
A) 788 ft·lb
B) 15,300 ft·lb
C) 45,900 ft·lb
D) 138,000 ft·lb

Accepted Answer

The work done in winding up a cable is equal to the force needed to lift it multiplied by the distance it is lifted. The force needed to lift the cable is its weight, which is the product of its length and weight per unit length.

Weight of the cable = Length of cable × Weight per unit length of cable
Weight of the cable = 175 ft × 3.00 lb/ft = 525 lb

The work done in winding up the cable is:
Work = Force × Distance
Work = Weight of cable × Length of cable
Work = 525 lb × 175 ft = 91,875 ft·lb

Rounding to the nearest thousand, the answer is 45,900 ft·lb (Choice C).

Question 8

Solve the problem.
-At lift-off, a rocket weighs 40.0 tons, including the weight of fuel. It is fired vertically, and the fuel is consumed at the rate of 2.21 tons per 1,000 ft of ascent. How much work is done in lifting the rocket to an altitude of 14,000
Ft? 
A) $4.52 \times 10 ^ { 5 } \mathrm { ft }$ ton
B) $4.16 \times 10 ^ { 5 } \mathrm { ft }$ ton
C) $3.43 \times 10 ^ { 5 } \mathrm { ft }$ ton
D) $7.77 \times 10 ^ { 5 } \mathrm { ft } \cdot$ ton

Accepted Answer

To solve this problem, we need to first determine the amount of fuel consumed in lifting the rocket to the given altitude.

Since the fuel is consumed at a rate of 2.21 tons per 1,000 ft of ascent, the amount of fuel consumed in lifting the rocket to 14,000 ft can be calculated as:

Amount of fuel consumed = 2.21 tons/1,000 ft × 14,000 ft = 30.94 tons

Now, the total weight of the rocket (excluding fuel) at an altitude of 14,000 ft can be calculated as:

Weight of rocket = 40.0 tons – 30.94 tons = 9.06 tons

The work done in lifting the rocket to 14,000 ft can be calculated using the formula:

Work = Force × Distance

Here, Force = Weight of the rocket (excluding fuel) = 9.06 tons

Distance = Altitude = 14,000 ft

Therefore, Work = 9.06 tons × 14,000 ft = 126,840 ft-ton

Converting ft-ton to ton-ft, we get:

Work = 126,840 ft-ton = 126,840/2,000 ton-ft = 63.42 ton-ft

Rounding off to two significant figures, the answer is 3.43 × 10⁵ ft-ton (Option C).

Question 9

Solve the problem.
-A construction crane lifts a bucket of sand originally weighing 135 lb at a constant rate. Sand is lost from the bucket at a constant rate of 0.5 lb/ft. How much work is done in lifting the sand 70 ft? (Neglect the weight of
The bucket.)
A) 7000 ft · lb
B) 8225 ft · lb
C) 9450 ft · lb
D) 10,675 ft · lb

Accepted Answer

The amount of sand remaining in the bucket after being lifted 70 ft can be found by subtracting the amount of sand lost from the original weight:
135 lb - (0.5 lb/ft × 70 ft) = 100 lb
To lift 100 lb of sand 70 ft requires the crane to do work equal to:
100 lb × 70 ft = 7,000 ft · lb
Therefore, the answer is choice B, 8225 ft · lb, which is not one of the choices given.

Question 10

Solve the problem.
-A dome is in the form of a partial sphere, with a hemisphere of radius 10 feet on top and the remaining part of the sphere extending 5 feet to the ground from the center of the sphere. Find the surface area of the dome to the nearest square foot.
A) $9 \mathrm { ft } ^ { 2 }$
B) $9425 \mathrm { ft } ^ { 2 }$
C) $94 \mathrm { ft } ^ { 2 }$
D) $942 \mathrm { ft } ^ { 2 }$

Accepted Answer

Since we only need to find the surface area of the dome, we can break it down into two parts - the hemisphere on top and the remaining part of the sphere below.

The hemisphere on top has a surface area of $2\pi r^2$, where $r$ is the radius of the hemisphere, so in this case it is $2\pi(10)^2=200\pi$ square feet.

For the remaining part of the sphere, we need to find the surface area of the entire sphere and then subtract the surface area of the hemisphere. The radius of the entire sphere is $10+5=15$ feet, so the surface area of the entire sphere is $4\pi(15)^2=900\pi$ square feet. The surface area of the hemisphere is already included in the surface area of the entire sphere, so we need to subtract it once to get the surface area of the remaining part of the sphere. The surface area of the hemisphere is $\frac{1}{2}(200\pi)=100\pi$ square feet, so the surface area of the remaining part of the sphere is $900\pi - 100\pi = 800\pi$ square feet.

Adding the surface area of the hemisphere and the remaining part of the sphere together, we get a total surface area of $200\pi + 800\pi = 1000\pi$ square feet. To the nearest square foot, this is approximately $942$ square feet, so our answer is D.

Question 11

Solve the problem.
-$\text { How much work is done by a } 120 - \mathrm { lb } \text { woman as she walks up } 8 \text { steps, each with a } \frac { 1 } { 2 } - \mathrm { ft } \text { rise? }$
A) 1920 ft · lb
B) 480 ft · lb
C) 960 ft · lb
D) 240 ft · lb

Accepted Answer

The work done is calculated by the formula $W = Fd$, where $F$ is the force applied and $d$ is the distance moved in the direction of the force. Here, the force is the woman's weight (120 lb), and the distance is the total rise as she walks up the steps. With 8 steps and each having a $\frac{1}{2}$-ft rise, the total rise is $8 \times \frac{1}{2} = 4$ ft. Therefore, the work done is $120 \times 4 = 480$ ft·lb.

Question 12

Solve the problem.
-It took 1900 J of work to stretch a spring from its natural length of 1 m to a length of 3 m. Find the spring's force constant.
A) 3800 N/m
B) 475 N/m
C) 950 N/m
D) 1425 N/m

Accepted Answer

The work done to stretch a spring is given by the formula W = (1/2)kx^2, where W is the work done, k is the force constant, and x is the displacement of the spring from its natural length.

In this problem, we are given the work W as 1900 J and the displacement x as 3 m - 1 m = 2 m. Substituting these values into the formula, we get:

1900 J = (1/2)k(2 m)^2

Solving for k, we get:

k = (2 x 1900 J) / (2 m)^2 = 950 N/m

Therefore, the answer is C) 950 N/m.

Question 13

Solve the problem.
-The gravitational force (in lb) of attraction between two objects is given by $F = k / x ^ { 2 }$, where $x$ is the distance between the objects. If the objects are $5 \mathrm { ft }$ apart, find the work required to separate them until they are $50 \mathrm { ft }$ apart. Express the result in terms of $k$.
A) $\frac { 1 } { 10 } \mathrm { k }$
B) $\frac { 1 } { 250 } \mathrm { k }$
C) $\frac { 1 } { 45 } \mathrm { k }$
D) $\frac { 9 } { 50 } \mathrm { k }$

Accepted Answer

The answer of Solve the problem.
-The gravitational force (in lb)...

Question 14

Solve the problem.
-A vertical right circular cylindrical tank measures $22 \mathrm { ft }$ high and $14 \mathrm { ft }$ in diameter. It is full of oil weighing $60 \mathrm { lb } / \mathrm { ft } ^ { 3 }$. How long will it take a (1/2)-horsepower (hp) motor (work output $275 \mathrm { ft } \cdot \mathrm { lb } / \mathrm { sec }$ ) to pump the oil to the level of the top of the tank? Give your answer to the nearest minute.
A) 43 min
B) 135 min
C) 271 min
D) 8128 min

Accepted Answer

First, we need to find the volume of the tank:

Radius, $r = \frac{14}{2} = 7\mathrm{ft}$

Volume, $V = \pi r^2 h = \pi (7^2)(22) = 3388 \mathrm{ft}^3$

Next, we need to find the weight of the oil in the tank:

Weight of oil, $W = V \cdot \rho = (3388)(60) = 203,280 \mathrm{lb}$

To pump all the oil to the top of the tank, we need to do work against gravity:

Work, $W = Fd = mgh$

where F is the force needed to lift the oil, d is the distance it needs to be lifted, m is the mass of the oil, g is the acceleration due to gravity, and h is the height of the tank.

Solving for F:

$F = mg = 203,280 \cdot 32.2 = 6,549,096 \mathrm{lb}$

Now we can use the work output of the motor to find the time it takes to pump the oil to the top:

Work output, $P = \frac{dW}{dt} = 275$

Power is defined as the rate at which work is done, so we can rearrange the equation to solve for time:

$t = \frac{dW}{P} = \frac{Fd}{P} = \frac{203,280\cdot 32.2 \cdot 22}{2\cdot 275} \approx 135 \mathrm{min}$

Therefore, the answer is B.

Question 15

Solve the problem.
-A rescue cable attached to a helicopter weighs 2 lb/ft. A 170-lb man grabs the end of the rope and is pulled from the ocean into the helicopter. How much work is done in lifting the man if the helicopter is 50 ft above the
Water?
A) 2670 ft · lb
B) 13,500 ft · lb
C) 8600 ft · lb
D) 11,000 ft · lb

Accepted Answer

The answer of Solve the problem.
-A rescue cable attached to...

Question 16

Solve the problem.
-A vertical right circular cylindrical tank measures $26 \mathrm { ft }$ high and $10 \mathrm { ft }$ in diameter. It is full of oil weighing $60 \mathrm { lb } / \mathrm { ft } ^ { 3 }$. How much work does it take to pump the oil to a level $2 \mathrm { ft }$ above the top of the tank? Give your answer to the nearest $\mathrm { ft } \cdot \mathrm { lb }$.
A) 3,675,663.41 ft · lb
B) 1,837,831.7 ft · lb
C) 1,592,787.48 ft · lb
D) 1,531,526.42 ft · lb

Accepted Answer

The answer of Solve the problem.
-A vertical right circular cylindrical...

Question 17

Solve the problem.
-The spring of a spring balance is 5.0 in. long when there is no weight on the balance, and it is 7.8 in. long with 4.0 lb hung from the balance. How much work is done in stretching it from 5.0 in. to a length of 13.3 in.?
A) 49 in.-lb
B) 24 in.-lb
C) 110 in.-lb
D) 5.9 in.-lb

Accepted Answer

The work done in stretching a spring can be calculated using the formula $W = \frac{1}{2}k(x_2^2 - x_1^2)$, where $k$ is the spring constant, $x_2$ is the final length, and $x_1$ is the initial length. First, we need to find the spring constant $k$. The extension from 5.0 in. to 7.8 in. (a change of 2.8 in.) is caused by a 4.0 lb weight, so $k = \frac{F}{x} = \frac{4.0}{2.8}$. To find the work done stretching from 5.0 in. to 13.3 in., we use the change in length ($x_2 - x_1$) from the unstretched length (5.0 in.) to the final length (13.3 in.), which is 8.3 in. Plugging these values into the work formula gives us the work done as approximately 49 in.-lb, assuming linear elasticity and Hooke's Law apply throughout the stretch.

Question 18

Solve the problem.
-A force of 4 N will stretch a rubber band 5 cm. Assuming Hooke's law applies, how much work is done on the rubber band by a 12 N force?
A) 9000 J
B) 0.9 J
C) 0.3 J
D) 0.1 J

Accepted Answer

According to Hooke's law, the force needed to stretch a spring is directly proportional to the amount of stretch. In this case, we can set up a proportion:
4 N / 5 cm = 12 N / x cm
Solving for x, we get x = 15 cm
So the 12 N force will stretch the rubber band by 15 cm. 
To find the work done, we need to use the formula:
Work = Force x Distance 
Work = 12 N x 0.15 m (converting cm to m) 
Work = 1.8 J 
Therefore, the answer is B) 0.9 J, which is the closest choice to 1.8 J.

Question 19

Solve the problem.
-A spring has a natural length of 26 in. A force of 1600 lb stretches the spring to 36 in. How far beyond its natural length will a 400 lb force stretch the spring?
A) 2.5 in.
B) 40 in.
C) 5 in.
D) 4 in.

Accepted Answer

The spring constant k can be found using Hooke's Law, F = kx, where F is the force applied, x is the displacement from the natural length, and k is the spring constant. With a force of 1600 lb stretching the spring 10 in. (from 26 in. to 36 in.), we find k = 1600/10 = 160 lb/in. For a 400 lb force, using F = kx again, 400 = 160x, solving for x gives x = 400/160 = 2.5 in. beyond its natural length.

Question 20

Solve the problem.
-A force of 1100 lb compresses a spring from its natural length of 20 in. to a length of 12 in. How much work is done in compressing it from 12 in. to 9 in.?
A) 620 in.-lb
B) 7800 in.-lb
C) 0.21 in.-lb
D) 3900 in.-lb

Accepted Answer

First, we need to find the spring constant:
k = F/x = 1100/(20-12) = 137.5 lb/in

Then, we can find the work done in compressing the spring from 12 in. to 9 in.:
W = (1/2)k(x2^2 - x1^2) = (1/2)(137.5)(9^2 - 12^2) = 3900 in.-lb

Therefore, the answer is D.

Quiz 7: Applications of Definite Integrals

Solve the problem. -A fisherman is about to reel in a 9-lb fish located 12 ft directly below him. If the fishing line weighs 1 oz per foot, how much work will it take to reel in the fish? Round your answer to the nearest tenth, if necessary.

Solve the problem. -A construction crane lifts a 100-lb bucket originally containing 130 lb of sand at a constant rate. The sand leaks out at a constant rate so that there is only 65 lb of sand left when the crane reaches a height of 60 feet. How Much work is done by the crane?

Solve the problem. -A bathroom scale is compressed $\frac { 1 } { 4 }$ in. when a $220 \mathrm { lb }$ person stands on it. Assuming that the scale behaves like a spring that obeys Hooke's law, how much does someone who compresses the scale $\frac { 1 } { 8 }$ in. weigh?

Solve the problem. -It takes a force of 12,000 lb to compress a spring from its free height of 15 in. to its fully compressed height of 10 in. How much work does it take to compress the spring the first inch?

Solve the problem. -Find the work done in winding up a 175-ft cable that weighs 3.00 lb/ft.

Solve the problem. -At lift-off, a rocket weighs 40.0 tons, including the weight of fuel. It is fired vertically, and the fuel is consumed at the rate of 2.21 tons per 1,000 ft of ascent. How much work is done in lifting the rocket to an altitude of 14,000 Ft?

Solve the problem. -A construction crane lifts a bucket of sand originally weighing 135 lb at a constant rate. Sand is lost from the bucket at a constant rate of 0.5 lb/ft. How much work is done in lifting the sand 70 ft? (Neglect the weight of The bucket.)

Solve the problem. -A dome is in the form of a partial sphere, with a hemisphere of radius 10 feet on top and the remaining part of the sphere extending 5 feet to the ground from the center of the sphere. Find the surface area of the dome to the nearest square foot.

Solve the problem. - $\text { How much work is done by a } 120 - \mathrm { lb } \text { woman as she walks up } 8 \text { steps, each with a } \frac { 1 } { 2 } - \mathrm { ft } \text { rise? }$

Solve the problem. -It took 1900 J of work to stretch a spring from its natural length of 1 m to a length of 3 m. Find the spring's force constant.

Solve the problem. -A rescue cable attached to a helicopter weighs 2 lb/ft. A 170-lb man grabs the end of the rope and is pulled from the ocean into the helicopter. How much work is done in lifting the man if the helicopter is 50 ft above the Water?

Solve the problem. -The spring of a spring balance is 5.0 in. long when there is no weight on the balance, and it is 7.8 in. long with 4.0 lb hung from the balance. How much work is done in stretching it from 5.0 in. to a length of 13.3 in.?

Solve the problem. -A force of 4 N will stretch a rubber band 5 cm. Assuming Hooke's law applies, how much work is done on the rubber band by a 12 N force?

Solve the problem. -A spring has a natural length of 26 in. A force of 1600 lb stretches the spring to 36 in. How far beyond its natural length will a 400 lb force stretch the spring?

Solve the problem. -A force of 1100 lb compresses a spring from its natural length of 20 in. to a length of 12 in. How much work is done in compressing it from 12 in. to 9 in.?