Question 1

Use a calculator to approximate the area of the surface generated when the given curve is revolved about the x-axis. Round to two decimal places when necessary.
-y = ln  on [1,  ]
A) 4.62
B) 15.18
C) 2.42
D) 29.05

Accepted Answer

The answer of Use a calculator to approximate the area...

Question 2

Solve the problem.   
-How much work is required to move an object from x = 0 to x = 6 (measured in meters) in the presence of a constant force of 10 N acting along the x-axis?
A) 2 J
B)$\infty$
C) 180 J
D) 60 J

Accepted Answer

The work done is calculated using the formula Work = Force x Distance. Here, the force is 10 N and the distance is 6 meters, so the work done is 10 N * 6 m = 60 Joules.

Question 3

Solve the problem.   
-A 20-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and that the chain has a density of 10 kg/m. How much work is required to wind the entire chain onto the cylinder using the winch?
A) 4000g J
B) 200g J
C) 2000g J
D) 3900g J

Accepted Answer

The work done to wind the chain is calculated by integrating the force needed to lift each infinitesimal segment of the chain over the entire length. The force at a point is the weight of the chain below that point. The total work is given by the integral of the weight function over the length of the chain, which results in 2000g J.

Question 4

Solve the problem.   
-A 50-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and that the chain has a density of 10 kg/m. How much work is required to wind the chain onto the cylinder if a 50-kg block is attached to the end of the chain?
A) 27,500g J
B) 25,050g J
C) 15,000g J
D) 12,550g J

Accepted Answer

The answer of Solve the problem.   
-A 50-m-long...

Question 5

Solve the problem.   
-A 60-m-long, 9.1-mm-diameter rope hangs free from a ledge. The density of the rope is 70 g/m. How much work is needed to pull the entire rope to the ledge?
A) 4.2g J
B) 9 g J
C) 126,000g J
D) 126 g J

Accepted Answer

The work needed to pull the rope to the ledge can be calculated by integrating the force needed to lift each segment of the rope from its initial position to the ledge. The force needed to lift a segment of the rope is equal to its weight, which depends on its mass and the acceleration due to gravity (g). The mass of a segment of the rope can be found by multiplying its length by the density of the rope.Given:- Length of the rope (L) = 60 m- Density of the rope (ρ) = 70 g/m- Diameter of the rope is not needed for this calculation.The mass of the entire rope (m) = ρ * L = 70 g/m * 60 m = 4200 gThe work done to lift a small segment of the rope (dm) from height x to the ledge is dW = g * x * dm, where g is the acceleration due to gravity. Since dm = ρ * dx, where dx is a small segment of the rope, the total work (W) needed to pull the entire rope to the ledge is the integral of dW from 0 to L.W = ∫ from 0 to L of g * x * ρ dx = g * ρ * ∫ from 0 to L of x dx = g * ρ * [x^2/2] from 0 to L = g * ρ * L^2 / 2Substituting the given values (and noting that g is factored into the final units as "g J" to indicate the gravitational constant is considered without specifying its numerical value):W = 70 g/m * (60 m)^2 / 2 = 126,000 g * m^2 / m = 126,000 g J (since g * m^2 / m simplifies to g J).However, considering the context of the question and the options provided, it seems there might have been a misunderstanding in the interpretation of units or the calculation process. The correct approach involves considering the gravitational force and integrating over the length of the rope, but the final answer should reflect the work done in a consistent unit of energy. Given the options and the common confusion around unit conversions, the intended answer seems to be "D) 126 g J," assuming a simplification or error in the calculation process presented in the options. This answer assumes a conceptual understanding of the problem rather than a strict adherence to the detailed calculation, which would not directly yield "126 g J" without additional context or assumptions about the units of g (gravity) being directly included in the answer choice.

Question 6

Solve the problem.   
-A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 45 N. How much work is done in stretching the spring 3 m from its equilibrium position?
A) 67.5 J
B) 405 J
C) 135 J
D) 810 J

Accepted Answer

The work done on a spring is given by the formula $W = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the displacement from the equilibrium position. First, we find $k$ using the given force and displacement: $F = kx$, so $k = \frac{F}{x} = \frac{45\,N}{0.5\,m} = 90\,N/m$. Then, we calculate the work for a 3 m displacement: $W = \frac{1}{2}(90\,N/m)(3\,m)^2 = \frac{1}{2}(90)(9) = 405\,J$.

Question 7

Solve the problem.   
-A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 45 N. How much work is done in compressing the spring 2.5 m from its equilibrium position?
A) 112.5 J
B) 56.25 J
C) 562.5 J
D) 281.25 J

Accepted Answer

The work done on a spring is given by the formula W = (1/2)kx^2, where k is the spring constant and x is the displacement. First, find k using F = kx: 45 N = k(0.5 m), so k = 90 N/m. Then, calculate the work for 2.5 m: W = (1/2)(90 N/m)(2.5 m)^2 = 281.25 J.

Question 8

Solve the problem.   
-A heavy-duty shock absorber is compressed 2 cm from its equilibrium position by a mass of 600 kg. How much work is required to compress the shock absorber 5 cm from its equilibrium position? (A mass of 600 kg exerts a force (in N) of 600g, where g $\approx$ 9.8 m/  .)
A) 367.5 J
B) 37.5 J
C) 3750 J
D) 36,750 J

Accepted Answer

The answer of Solve the problem.   
-A heavy-duty...

Question 9

Solve the problem.   
-It takes 100 J of work to stretch a spring 1 m from its equilibrium position. How much work is needed to stretch it an additional 1 m?
A) 2500 J
B) 1500 J
C) 300 J
D) 500 J

Accepted Answer

The work needed to stretch a spring is proportional to the square of the displacement from its equilibrium position, according to Hooke's law. If 100 J is needed for the first meter, stretching it to 2 meters (twice the original distance) requires four times the work (since 2^2 = 4), making it 400 J in total. The additional work needed for the second meter is the difference: 400 J - 100 J = 300 J.

Question 10

Solve the problem.   
-Hooke's law is applicable to idealized (linear) springs that are not stretched or compressed too far. Consider a nonlinear spring whose restoring force is given by F(x) = 18x - 0.3  , for |x| $\le$ 9. How much work is done in stretching the spring from its equilibrium position (x = 0) to x = 1? Round to two decimal places when appropriate.
A) 473.85 J
B) 8.93 J
C) 236.93 J
D) 17.7 J

Accepted Answer

The answer of Solve the problem.   
-Hooke's law...

Question 11

Solve the problem.   
-Hooke's law is applicable to idealized (linear) springs that are not stretched or compressed too far. Consider a nonlinear spring whose restoring force is given by F(x) = 15x - 0.3  , for |x| $\le$ 8. How much work is done in compressing the spring from its equilibrium position (x = 0) to x = -0.5? Round to two decimal places when appropriate.
A) 1.87 J
B) 172.8 J
C) 345.6 J
D) 3.73 J

Accepted Answer

The answer of Solve the problem.   
-Hooke's law...

Question 12

Solve the problem.   
-A 25-kg mass is attached to a spring that hangs vertically and is stretched 3 m from the equilibrium position of the spring. Assume a linear spring with F(x) = kx. How much work is required to compress the spring and lift the mass 1.5 m? Round to four decimal places when appropriate.
A) 275.625 J
B) 398.125 J
C) 30.625 J
D) 643.125 J

Accepted Answer

The answer of Solve the problem.   
-A 25-kg...

Question 13

Solve the problem.   
-A 25-kg mass is attached to a spring that hangs vertically and is stretched 1 m from the equilibrium position of the spring. Assume a linear spring with F(x) = kx. How much work is required to stretch the spring and lower the mass 1.5 m? Round to four decimal places when appropriate.
A) 459.375 J
B) 1010.625 J
C) 91.875 J
D) 643.125 J

Accepted Answer

The answer of Solve the problem.   
-A 25-kg...

Question 14

Solve the problem.   
-A swimming pool has the shape of a box with a base that measures 29 m by 15 m and a depth of 3 m. How much work is required to pump the water out of the pool when it is full?
A) 1,957,500 J
B) 19,183,500 J
C) 38,367,000 J
D) 72,765,000 J

Accepted Answer

The work required to pump the water out of the pool is equal to the weight of the water multiplied by the height to which it must be pumped. The weight of the water is equal to its density multiplied by its volume. The volume of the water is equal to the area of the base of the pool multiplied by its depth. The height to which the water must be pumped is equal to the depth of the pool. Therefore, the work required to pump the water out of the pool is:W = ρVghW = (1000 kg/m^3)(29 m x 15 m x 3 m)(9.8 m/s^2)W = 19,183,500 J

Question 15

Solve the problem.   
-A cylindrical water tank has height 12 m and radius 2 m (see figure). If the tank is full of water, how much work is required to pump the water to the level of the top of the tank and out of the tank? Express the answer in terms of $\pi$. 
A) 33,868,800? J
B) 705,600 J
C) 2,822,400$\pi$J
D) 784,000$\pi$ J

Accepted Answer

The answer of Solve the problem.   
-A cylindrical...

Question 16

Solve the problem.   
-A water tank is shaped like an inverted cone with height 18 m and base radius 3 m (see figure). If the tank is full, how much work is required to pump the water to the level of the top of the tank and out of the tank? Express the answer in terms of $\pi$.

A) 2,381,400$\pi$ J
B) 1,587,600 J
C) 14,288,400$\pi$ J
D) 85,730,400$\pi$ J

Accepted Answer

The answer of Solve the problem.   
-A water...

Question 17

Solve the problem.   
-A swimming pool is 16 m long and 10 m wide, with a bottom that slopes uniformly from a depth of  at one end to a depth of 2 m at the other end (see figure). Assuming the pool is full, how much work is required to pump the water to a level 0.2 m above the top of the pool? Round to one decimal place when appropriate.

A) 1,097,600 J
B) 9,042,133.3 J
C) 17,404,800 J
D) 7,944,533.3 J

Accepted Answer

The answer of Solve the problem.   
-A swimming...

Question 18

Solve the problem.   
-A spherical water tank with an inner radius of 4 m has its lowest point 3 m above the ground. It is filled by a pipe that feeds the tank at its lowest point (see figure). Neglecting the volume of the inflow pipe, how much work is required to fill the tank if it is initially empty? Express the answer in terms of $\pi$.

A) 5,017,600/ 3 $\pi$ J
B) 2,508,800 /3 $\pi$ J
C) 627,200/3 $\pi$ J
D) 1,254,400/3 $\pi$ J

Accepted Answer

The answer of Solve the problem.   
-A spherical...

Question 19

Solve the problem.   
-A water trough has a semicircular cross section with a radius of 0.5 m and a length of 4 m (see figure). How much work is required to pump water out of the trough when it is full? Round to two decimal places when appropriate. 
A) 1633.33 J
B) 3266.67 J
C) 19,600 J
D) 38,587.5$\pi$ J

Accepted Answer

The answer of Solve the problem.   
-A water...

Question 20

Solve the problem.   
-A glass has circular cross sections that taper (linearly) from a radius of 7 cm at the top of the glass to a radius of 6 cm at the bottom. The glass is 15 cm high and full of lemonade. How much work is required to drink all the lemonade through a straw if your mouth is 8 cm above the top of the glass? Assume the density of lemonade equals the density of water. Round to two decimal places when appropriate.
A) 2.96 J
B) 0.94 J
C) 55.65 J
D) 4.32 J

Accepted Answer

The work done is equal to the change in potential energy of the lemonade. The potential energy of the lemonade is equal to the mass of the lemonade times the acceleration due to gravity times the height of the lemonade. The mass of the lemonade is equal to the density of the lemonade times the volume of the lemonade. The volume of the lemonade is equal to the area of the circular cross section of the glass times the height of the lemonade. The area of the circular cross section of the glass is equal to pi times the radius of the circular cross section squared. The height of the lemonade is equal to the height of the glass minus the height of the straw. The acceleration due to gravity is 9.8 m/s^2.Plugging in the given values, we get:mass of the lemonade = (1000 kg/m^3) * (pi * (0.07 m)^2 * 0.15 m) = 0.226 kgheight of the lemonade = 0.15 m - 0.08 m = 0.07 mwork done = (0.226 kg) * (9.8 m/s^2) * (0.07 m) = 0.15 JTherefore, the work required to drink all the lemonade through a straw is 0.15 J.

Quiz 6: Applications of Integration

Use a calculator to approximate the area of the surface generated when the given curve is revolved about the x-axis. Round to two decimal places when necessary. -y = ln on [1, ]

Solve the problem. -How much work is required to move an object from x = 0 to x = 6 (measured in meters) in the presence of a constant force of 10 N acting along the x-axis?

Solve the problem. -A 20-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and that the chain has a density of 10 kg/m. How much work is required to wind the entire chain onto the cylinder using the winch?

Solve the problem. -A 50-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and that the chain has a density of 10 kg/m. How much work is required to wind the chain onto the cylinder if a 50-kg block is attached to the end of the chain?

Solve the problem. -A 60-m-long, 9.1-mm-diameter rope hangs free from a ledge. The density of the rope is 70 g/m. How much work is needed to pull the entire rope to the ledge?

Solve the problem. -A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 45 N. How much work is done in stretching the spring 3 m from its equilibrium position?

Solve the problem. -A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 45 N. How much work is done in compressing the spring 2.5 m from its equilibrium position?

Solve the problem. -A heavy-duty shock absorber is compressed 2 cm from its equilibrium position by a mass of 600 kg. How much work is required to compress the shock absorber 5 cm from its equilibrium position? (A mass of 600 kg exerts a force (in N) of 600g, where g ≈\approx≈ 9.8 m/ .)

Solve the problem. -It takes 100 J of work to stretch a spring 1 m from its equilibrium position. How much work is needed to stretch it an additional 1 m?

Solve the problem. -A swimming pool has the shape of a box with a base that measures 29 m by 15 m and a depth of 3 m. How much work is required to pump the water out of the pool when it is full?

Solve the problem. -A cylindrical water tank has height 12 m and radius 2 m (see figure) . If the tank is full of water, how much work is required to pump the water to the level of the top of the tank and out of the tank? Express the answer in terms of π\piπ .

Solve the problem. -A water tank is shaped like an inverted cone with height 18 m and base radius 3 m (see figure) . If the tank is full, how much work is required to pump the water to the level of the top of the tank and out of the tank? Express the answer in terms of π\piπ .

Solve the problem. -A water trough has a semicircular cross section with a radius of 0.5 m and a length of 4 m (see figure) . How much work is required to pump water out of the trough when it is full? Round to two decimal places when appropriate.