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Work on a Block Sliding Up a Frictionless Incline:

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Block weight 15.0 N sits on a frictionless inclined plane, which makes an angle θ = 23.0° with respect to the horizontal, as shown in the figure. A force of magnitude F = 5.86 N, applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed.

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a) The block moves up an incline with constant speed. What is the total work W. Total done on the block by all forces as the block moves a distance L = 3.40 m up the incline? Include only the work done after the block has started moving at constant speed, not the work needed to start the block moving from rest.

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Since the block is moving at a constant speed there is no change in kinetic energy. Since work is the change in kinetic energy and there is no change, the net work on the block is zero.

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b) What is Wg, the work done on the block by the force of gravity w‖ as the block moves a distance L = 4.50m up the incline?

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1) Find the component of the gravitational force parallel to the plane
What is w||, the magnitude of the component of the force of gravity along the inclined plane? →w parallel to the inclined plane has magnitude given by w‖=wsinθ. = 40*sin(22) = 15.
2) Wg = -mgy = -15*4.5 = -67J

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What is Wƒ, the work done on the block by the applied force F→ as the block moves a distance = 4.5m up the incline?

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Since we found the net work to be zero (in Part A), the work done by the applied force has to offset the work done by gravity: Wg=−Wƒ. ∴ 67J

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The mechanical energy of a system is defined as the sum of kinetic energy K and potential energy U. For such systems in which no forces other than the gravitational and elastic forces do work, the law of conservation of energy can be written as

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Ki+Ui=Kf+Uf,
K=½mv²
Potential: Ug = mgh
Elastic potential energy: For a spring with a force constant k, stretched or compressed a distance x, the associated elastic potential energy is
Ue=½kx²

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b) Which form of the law of conservation of energy describes the motion of the block when it slides from the top of the table to the bottom of the ramp?

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Think about these questions:
Are there any nonconservative forces acting on the block during this part of the trip?
Are there any objects involved that can store elastic potential energy?
Is the block changing its height?
Is the block changing its speed?
½mv₁² + mgh₁ = ½mv₂² + mgh₂

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c) What if nonconservative forces, such as friction, also act within the system? In that case, the total mechanical energy will change. The law of conservation of energy is then written as

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½mv₁² + mgh₁ + ½kx₁² + Wnc = ½mv₂² + mgh₂ + ½kx₂²
Where Wnc represents the work done by the nonconservative forces acting on the object between the initial and the final moments. The work Wnc is usually negative; that is, the nonconservative forces tend to decrease, or dissipate, the mechanical energy of the system.

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d) Using conservation of energy, find the speed vb of the block at the bottom of the ramp.
Express your answer in terms of some or all the variables m, v, h and and any appropriate constants.

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PE₀ + KE₀ = PEƒ + KEƒ
At the top of the ramp, potential energy will be at a maximum and kinetic energy will be at a minimum. At the bottom of the ramp, kinetic energy will be at a maximum and potential energy will be minimum. But although the type of energy changes, the total amount of energy remains the same:
mgh + ½mv² = mgh(b) + ½mv(b)²
Since the height at the bottom is zero, we can eliminate potential energy from the right side of the formula:
mgh + ½mv² = ½mv(b)²
gh + ½v² = ½v(b)²
2gh + v² = v(b)²
v(b) = √(v² + 2gh)
== √(v² + 2gh)

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As the block slides across the floor, what happens to its kinetic energy K, potential energy U, and total mechanical energy E?

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K decreases
U stays the same
E decreases

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g) What force is responsible for the decrease in the MECHANICAL ENERGY of the block?

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friction

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Mechanical energy of a system

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Sum of kinetic energy K and potential energy U. For such systems in which no forces other than the gravitational and elastic forces do work, the law of conservation of energy can be written as

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A block of mass m slides at a speed v along a horizontal smooth table. It next slides down a smooth ramp, descending a height h, and then slides along a horizontal rough floor, stopping eventually. Assume that the block slides slowly enough so that it does not lose contact with the supporting surfaces (table, ramp, or floor).
You will analyze the motion of the block at different moments using the law of conservation of energy.
a) Which word in the statement of this problem allows you to assume that the table is frictionless?

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Although there are no truly "frictionless" surfaces, sometimes friction is small enough to be neglected. The word "smooth" often describes such low-friction surfaces. Can you deduce what the word "rough" means?

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h) Find the amount of energy E (MECH ENEG) dissipated by friction by the time the block stops. Express your answer in terms of some or all the variables m, v, and h and any appropriate constants.

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Friction will slow the block to zero, so it will dissipate all of the kinetic and potential energy that the block has at the top of the ramp.
= ½mv² + mgh

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5.33: Find the work done by friction. (SI: J)
A 1.37kg book slides 1.26m along a level surface. The coefficient of kinetic friction between book and surface is 0.154.

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work done = force x distance
mg = 1.37kg x 9.8 = 13.426N
13.426 x 0.154 = force required to overcome friction = 2.07N (rounded)
(0 - 2.07) = -2.07 N (force of friction)
-2.07 x 1.26
≅ -2.61J (answer)

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Spider Silk: Spider silk is one of the most remarkable elastic materials known. Consider a silk strand suspended vertically with a 0.35-g fly stuck on the end. With the fly attached, the silk measures 28.0 cm in length. The resident spider, of mass 0.66 g, senses the fly and climbs down the silk to investigate. With both spider and fly at the bottom, the silk measures 37.5 cm

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So a mass of 0.66g caused a change in length of (37.5 - 28.0) = 9.5 cm
(9.5 ÷ 100) = 0.095 m [to match ƒg units]

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a) Find the spring constant of the silk.

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So a mass of 0.66g caused length Δ of (37.5 - 28.0) = 9.5 cm ↔ 0.095m {gravity is in meters}
Recall that the spring constant is normally expressed in Newtons per meter
(ƒg =9.8 and 0.66 g = 0.00066 kg )
0.00066 * 9.8 = 0.006468 N,
Spring constant = (0.006468 ÷ 0.095 Nm⁻¹) = 0.068 Nm⁻¹
≡ 6.8 * 10⁻² (N/m)

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b) Find the equilibrium length of the silk.

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Fly's weight in N = 0.00035 * 9.8
= 0.00343 N
∴ amount the fly stretches the thread must be: Mass(N) ÷ Spring Constant
= 0.00343 ÷ 0.068 = 0.050 m
= 5 cm
Equilibrium length = 28 cm- 5cm
≅ 23 cm