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Energy and work problems

A net force of 10 N moves a 3 kg object initially at rest a horizontal distance of 5 m directly in the direction the net force is acting. Hint: The first part of Example Problem 3 is similar to this problem.

What is the velocity (in m/s) of the object when the force has moved it 5 m?
A rope is tied to a box and used to pull the box 1.5 m along a horizontal floor. The rope makes an angle of 30 degrees with the horizontal and has a tension of 5 N. The opposing friction force between the box and the floor is 1 N.

How much work does gravity do on the box?
How much work does the tension do on the box?
How much work does friction do on the box?
In this process, what is the work done on the box?
Tendons are strong elastic fibers that attach muscles to bones. To a reasonable approximation, they obey Hookes law. In laboratory tests on a particular tendon, it was found that, when a 250-g object was hung from it, the tendon stretched 1.23 cm. Because of its thickness, the maximum tension this tendon can support without rupturing is 138 N.

What is the maximum length the tendon can stretch without rupturing?
How much energy is stored in the tendon at that max length?
You overindulged in a delicious dessert, so you plan to work off the extra calories at the gym. To accomplish this, you decide to do a series of arm curls holding a 5.00 kg weight in one hand. The distance from your elbow to the weight is 35.0 cm , and in each arm curl you start with arm straight down and raise the weight in your hand by pivoting at your elbow until your elbow is completely bent and the weight in your hand is vertical. Assume that the weight of your arm is small enough compared with the weight you are lifting that you can ignore it. As is typical, your muscles are 20.0% efficient in converting food energy into mechanical energy, with the rest going into heat.

If your dessert contained 350 food calories, how many arm curls must you do to work off these calories?
While hovering, a typical flying insect applies an average force equal to twice its weight during each downward stroke. Take the mass of the insect to be 10 g, and assume the wings move an average downward distance of 1.0 cm during each stroke.

Assuming 100 downward strokes per second, what is the average power output of the insect?
A 10 kg block at rest at a height of 7 m above the ground slides down an incline to the ground. Assume there is no friction between the block and the incline, and assume the coefficient of friction between the block and the ground is = 0.12. Hint: For help on this problem, watch the Example Problem 2 videos.

What is the velocity (in m/s) of the block at the bottom of the incline?
How far in meters does the block slide on the ground before it comes to rest? (See video “Example Problem 2 Part 4”)
A 10.2 kg weight hangs from a pulley as shown below, and an upward force on the rope (represented by the red arrow) raises the weight a distance of 8 cm (0.08 m) at a constant velocity. Image below by Csar Rincn.phpzJiTAk.png

What is the change in potential energy (in J) of the block in being lifted?
What is the upward force (in N) represented by the red arrow? (Hint: Look at the solutions for similar Newton’s Laws HW.)
How far (in m) is the rope pulled upwards? (Hint: Use work-energy concepts to solve this problem, i.e. work done by rope = change in energy of the block.)
What is the work (in J) done by the upward force in lifting the block?
What is the mechanical advantage of the pulley? (Hint: Watch “Hewitt-Drew-it! PHYSICS 33. Machines and Energy” video for a definition of Mechanical Advantage and an analysis of pulleys and other simple machines, or see your textbook.)

Homework Content
Question 1Question 11 PointA net force of 10 N moves a 3 kg object initially at rest a horizontal distance of 5 m directly in the direction the force is acting.What is the velocity in m/s of the object when the force has moved it 5 m?Integer, decimal, or E notation allowed
Question 2Question 21 PointA rope is tied to a box and used to pull the box 1.5 m along a horizontal floor. The rope makes an angle of 30 degrees with the horizontal and has a tension of 5 N. The opposing friction force between the box and the floor is 1 N.How much work does gravity do on the box?Integer, decimal, or E notation allowed
Question 3Question 31 PointA rope is tied to a box and used to pull the box 1.5 m along a horizontal floor. The rope makes an angle of 30 degrees with the horizontal and has a tension of 5 N. The opposing friction force between the box and the floor is 1 N.In this process, what is the work done on the box?Integer, decimal, or E notation allowed
Question 4Question 41 PointTendons are strong elastic fibers that attach muscles to bones. To a reasonable approximation, they obey Hookes law. In laboratory tests on a particular tendon, it was found that, when a 250-g object was hung from it, the tendon stretched 1.23 cm. Because of its thickness, the maximum tension this tendon can support without rupturing is 138 N.How much energy is stored in the tendon at the maximum length?Integer, decimal, or E notation allowed
Question 5Question 51 PointYou overindulged in a delicious dessert, so you plan to work off the extra calories at the gym. To accomplish this, you decide to do a series of arm curls holding a 5.00 kg weight in one hand. The distance from your elbow to the weight is 35.0 cm , and in each arm curl you start with arm straight down and raise the weight in your hand by pivoting at your elbow until your elbow is completely bent and the weight in your hand is vertical. Assume that the weight of your arm is small enough compared with the weight you are lifting that you can ignore it. As is typical, your muscles are 20.0% efficient in converting food energy into mechanical energy, with the rest going into heat.If your dessert contained 350 food calories, how many arm curls must you do to work off these calories?Integer, decimal, or E notation allowed
Question 6Question 61 PointWhile hovering, a typical flying insect applies an average force equal to twice its weight during each downward stroke. Take the mass of the insect to be 10 g, and assume the wings move an average downward distance of 1.0 cm during each stroke.Assuming 100 downward strokes per second, what is the average power output (in W) of the insect?Integer, decimal, or E notation allowed
Question 7Question 71 PointA 10 kg block initially at rest at a height of 7 m above the ground slides down an incline to the ground. Assume there is no friction between the block and the incline, and assume the coefficient of friction between the block and the ground is = 0.12.What is the velocity in m/s of the block at the bottom of the incline?Integer, decimal, or E notation allowed
Question 8Question 81 PointA 10 kg block initially at rest at a height of 7 m above the ground slides down an incline to the ground. Assume there is no friction between the block and the incline, and assume the coefficient of friction between the block and the ground is = 0.12.How far in meters does the block slide on the ground before it comes to rest?Integer, decimal, or E notation allowed
Question 9Question 91 PointA 10.2 kg weight hangs from a pulley as shown below, and an upward force on the rope (represented by the red arrow) raises the weight 8 cm (0.08 m). Image below by Csar Rincn.rincon weight on pulley.png
What is the change in potential energy in J of the block in being lifted?Integer, decimal, or E notation allowed
Question 10Question 101 PointA 10.2 kg weight hangs from a pulley as shown below, and an upward force on the rope (represented by the red arrow) raises the weight 8 cm (0.08 m). Image below by Csar Rincn.rincon weight on pulley.png
What is the upward force in N represented by the red arrow?Integer, decimal, or E notation allowed
Question 11Question 111 PointA 10.2 kg weight hangs from a pulley as shown below, and an upward force on the rope (represented by the red arrow) raises the weight 8 cm (0.08 m). Image below by Csar Rincn.rincon weight on pulley.png
How far is the rope pulled upwards?Integer, decimal, or E notation allowed
Question 12Question 121 PointA 10.2 kg weight hangs from a pulley as shown below, and an upward force on the rope (represented by the red arrow) raises the weight 8 cm (0.08 m). Image below by Csar Rincn.rincon weight on pulley.png
What is the work (in J) done by the upward force in lifting the block?Integer, decimal, or E notation allowed
Question 13Question 131 PointA 10.2 kg weight hangs from a pulley as shown below, and an upward force on the rope (represented by the red arrow) raises the weight 8 cm (0.08 m). Image below by Csar Rincn.rincon weight on pulley.png
What is the mechanical advantage of the pulley?

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Sample Answer

 

 

 

 

Here are the solutions to the physics problems:

Question 1

  • Given:

    • Force (F) = 10 N

    • Mass (m) = 3 kg

    • Initial velocity (v₀) = 0 m/s

    • Displacement (Δx) = 5 m

  • Find: Final velocity (v)

  • Solution:

    1. Use the work-energy theorem: Work done by the net force = change in kinetic energy.

    2. Work done by the net force: W = F * Δx = 10 N * 5 m = 50 J

    3. Change in kinetic energy: ΔKE = (1/2)mv² – (1/2)mv₀²

    4. Since initial velocity is 0: ΔKE = (1/2)mv²

    5. Equate the work and change in kinetic energy: 50 J = (1/2) * 3 kg * v²

    6. Solve for v: v = √(2 * 50 J / 3 kg) = 5.77 m/s

Question 2

  • Given:

    • Displacement (Δx) = 1.5 m

    • Force of gravity (Fg) = mg (where g = 9.8 m/s²)

  • Find: Work done by gravity (Wg)

  • Solution:

    1. Work done by gravity: Wg = Fg * Δx * cos θ

    2. Since gravity acts downwards and the displacement is horizontal, the angle θ = 90 degrees.

    3. Therefore, cos θ = 0, and Wg = 0 J.

Question 3

  • Given:

    • Force of tension (Ft) = 5 N

    • Angle of rope (θ) = 30 degrees

    • Displacement (Δx) = 1.5 m

    • Force of friction (Ff) = 1 N

  • Find: Work done on the box (W)

Full Answer Section

 

 

 

 

 

  • Solution:

    1. Work done by tension: Wt = Ft * Δx * cos θ = 5 N * 1.5 m * cos 30° = 6.5 J

    2. Work done by friction: Wf = Ff * Δx * cos 180° = -1 N * 1.5 m * (-1) = -1.5 J (negative since friction opposes motion)

    3. Total work done on the box: W = Wt + Wf = 6.5 J – 1.5 J = 5 J

Question 4

  • Given:

    • Mass of object (m) = 250 g = 0.25 kg

    • Extension (Δx) = 1.23 cm = 0.0123 m

    • Maximum tension (Tmax) = 138 N

  • Find:

    • Maximum length (Lmax)

    • Energy stored (U) at maximum length

  • Solution:

    1. Find the spring constant (k) using Hooke’s Law: T = kΔx

      • k = T/Δx = (0.25 kg * 9.8 m/s²) / 0.0123 m = 198.37 N/m

    2. Find the maximum extension (Δxmax) using Hooke’s Law and maximum tension:

      • Δxmax = Tmax / k = 138 N / 198.37 N/m = 0.696 m

    3. Calculate the maximum length: Lmax = initial length + Δxmax

      • Since we don’t have the initial length, we can only find the change in length, which is 0.696 m

    4. Calculate the energy stored: U = (1/2)k(Δxmax)²

      • U = (1/2) * 198.37 N/m * (0.696 m)² = 48.2 J

Question 5

  • Given:

    • Weight (m) = 5 kg

    • Distance (Δh) = 35 cm = 0.35 m

    • Efficiency (η) = 20% = 0.2

    • Food calories (E) = 350 calories

  • Find: Number of arm curls (n)

  • Solution:

    1. Convert food calories to Joules: 350 calories * 4184 J/calorie = 1,464,400 J

    2. Work done per arm curl: W = mgh = 5 kg * 9.8 m/s² * 0.35 m = 17.15 J

    3. Mechanical work done per arm curl (considering efficiency): W_mech = η * W = 0.2 * 17.15 J = 3.43 J

    4. Number of arm curls: n = E / W_mech = 1,464,400 J / 3.43 J/curl ≈ 426,443 curls

Question 6

  • Given:

    • Mass of insect (m) = 10 g = 0.01 kg

    • Force (F) = 2 * weight = 2 * mg

    • Distance (Δy) = 1 cm = 0.01 m

    • Strokes per second (f) = 100

  • Find: Average power (P)

  • Solution:

    1. Calculate the force: F = 2 * 0.01 kg * 9.8 m/s² = 0.196 N

    2. Work done per stroke: W = F * Δy = 0.196 N * 0.01 m = 0.00196 J

    3. Power: P = Work/time = W * f = 0.00196 J/stroke * 100 strokes/s = 0.196 W

Question 7

  • Given:

    • Mass (m) = 10 kg

    • Initial height (h₁) = 7 m

    • Final height (h₂) = 0 m

    • Coefficient of friction (μ) = 0.12

  • Find: Velocity (v) at the bottom of the incline

  • Solution:

    1. Use conservation of energy. Since there’s no friction on the incline, the initial potential energy is converted to kinetic energy at the bottom.

    2. Initial potential energy: PE₁ = mgh₁ = 10 kg * 9.8 m/s² * 7 m = 686 J

    3. Final kinetic energy: KE₂ = (1/2)mv²

    4. Equate PE₁ and KE₂: 686 J = (1/2) * 10 kg * v²

    5. Solve for v: v = √(2 * 686 J / 10 kg) = 11.71 m/s

Question 8

  • Given:

    • Velocity at the bottom of the incline (v) = 11.71 m/s

    • Coefficient of friction (μ) = 0.12

  • Find: Distance (Δx) the block slides before stopping

  • Solution:

    1. Use the work-energy theorem: Work done by friction = change in kinetic energy.

    2. Work done by friction: Wf = -μmgΔx (negative since friction opposes motion)

    3. Change in kinetic energy: ΔKE = 0 – (1/2)mv²

    4. Equate the work and change in kinetic energy: -μmgΔx = -(1/2)mv²

    5. Solve for Δx: Δx = (v²)/(2μg) = (11.71 m/s)² / (2 * 0.12 * 9.8 m/s²) = 5.86 m

Question 9

  • Given:

    • Mass (m) = 10.2 kg

    • Height change (Δh) = 8 cm = 0.08 m

  • Find: Change in potential energy (ΔPE)

  • Solution:

    • ΔPE = mgΔh = 10.2 kg * 9.8 m/s² * 0.08 m = 8.02 J

Question 10

  • Given:

    • Mass (m) = 10.2 kg

    • Height change (Δh) = 8 cm = 0.08 m

  • Find: Upward force (F)

  • Solution:

    • Since the block is lifted at constant velocity, the net force is zero. This means the upward force is equal in magnitude to the weight of the block.

    • F = mg = 10.2 kg * 9.8 m/s² = 100 N

Question 11

  • Given:

    • Work done by upward force (W) = 8.02 J (from Question 9)

    • Upward force (F) = 100 N (from Question 10)

  • Find: Distance the rope is pulled (Δy)

  • Solution:

    • W = F * Δy

    • Δy = W/F = 8.02 J / 100 N = 0.0802 m

Question 12

  • Given:

    • Upward force (F) = 100 N (from Question 10)

    • Distance the rope is pulled (Δy) = 0.0802 m (from Question 11)

  • Find: Work done by the upward force (W)

  • Solution:

    • W = F * Δy = 100 N * 0.0802 m = 8.02 J

Question 13

  • Mechanical Advantage: The mechanical advantage of a simple machine is the ratio of the output force to the input force. In this case, the output force is the weight of the block being lifted (100 N), and the input force is the force applied to the rope (also 100 N).

  • Mechanical Advantage of the pulley:

    • Mechanical Advantage (MA) = Output Force / Input Force = 100 N / 100 N = 1

This question has been answered.

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