LEARN MOREREMARKS Friction or drag from immersion in a fluid damps the motion of an object attached to aspring, eventually bringing the object to rest.QUESTION In the case of friction, what percent of the mechanical energy was lost by the time the massfirst reached the equilibrium point? (Hint: use the answers to parts (a) and (c).)%PRACTICE ITUse the worked example above to help you solve this problem. A block with mass of 5.79 kg is attached toa horizontal spring with spring constant k 3.32 x 10² N/m, as shown in the figure. The surface the blockrests upon is frictionless. The block is pulled out to x; = 0.0510 m and released.=(a) Find the speed of the block at the equilibrium point.m/s(b) Find the speed when x = 0.029 m.m/s(c) Repeat part (a) if friction acts on the block, with coefficient μk=m/s0.120. Substitute expressions for the block'skinetic energy and the potentialenergy, and set the gravity terms tozero.Substitute v; = 0, xf = 0, and multiplyby 2/m.Solve forvalues.VfSolve forvalues.and substitute the givenVfand substitute the given(B) Find the speed of the block at the halfway point.Set v₁ = 0 in Equation (1) and multiply2kx²mby 2/m.(C) Repeat part (a), this time with friction.Apply the work-energy theorem. Thework done by the force of gravity andthe normal force is zero because theseforces are perpendicular to the motion.0, xf = 0, and(1) 1/2mv ² + 1/{kx² = 1/2mv7 + = = kx7²Substitute v;W fric = μknx;.Set n = mg and solve for vf.Kx²=v²= √k xm= 0.447 m/sVf = V=-X¡ == Vv² +kx²mv₁ = √√ x²(x²-x²)Vf V4.00 × 10² N/m (0.0500 m)5.00 kgm4.00 × 10² N/m [(0.050 m)² – (0.025 m)²]5.00 kg0.387 m/sWfric = 1/2mv? - 1/2mv² + ½kx² - 1/1/kx7²- μxNx₁ = = = mv² - 1⁄2kx²1/2mv² = 1kx2² - μxmgx₁MkmgxiV₁ = √√ K x₁² - 2µkgx₁m4.00 × 10² N/m (0.050 m)² — 2(0.150)(9.80 m/s²)(0.050 m)5.00 kg= 0.230 m/s

Answered: Image /qna-images/answer/3471bac2-bd8f-49e6-bb74-08ef152a0edd.jpg

Use the worked example above to help you solve this problem. A block with mass of 5.79 kg is attached to a horizontal spring with spring constant k = 3.32 x 10² N/m, as shown in the figure. The surface the block rests upon is frictionless. The block is pulled out to x; = 0.0510 m and released. (a) Find the speed of the block at the equilibrium point. m/s (b) Find the speed when x = 0.029 m. m/s (c) Repeat part (a) if friction acts on the block, with coefficient μ = 0.120.

Use the worked example above to help you solve this problem. A block with mass of 5.79 kg is attached to a horizontal spring with spring constant k = 3.32 x 10² N/m, as shown in the figure. The surface the block rests upon is frictionless. The block is pulled out to x; = 0.0510 m and released. (a) Find the speed of the block at the equilibrium point. m/s (b) Find the speed when x = 0.029 m. m/s (c) Repeat part (a) if friction acts on the block, with coefficient μ = 0.120.

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter5: Analysis Of Convection Heat Transfer

Section: Chapter Questions

Problem 5.5P: Evaluate the dimensionless groups hcD/k,UD/, and cp/k for water, n-butyl alcohol, mercury, hydrogen,...

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