WileyPLUS: MyWileyPLUS I Help Contact Us I Log Ou Halliday, Fundamentals of Phys

Question

WileyPLUS: MyWileyPLUS I Help Contact Us I Log Ou Halliday, Fundamentals of Physics, 10e GENERAL PHYSICS 1 - 2 (PHY 201-202 Assignment Gradebook ORION Downloadable eTextbook ?? ent MESSAGE MY INSTRUCTOR FULL SCREEN PRINTER VERSION BACK NEXT Chapter 08, Problem 042 A worker pushed a 38.0 kg block 10.0 m along a level floor at constant speed with a force directed 21.0° below the horizontal. If the coefficient of kinetic friction between block and floor was 0.400, what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system? (a) Number (b) Number Show Work is REQUIRED for this question: Units Units Qpen Show Work SHOW HINT LINK TO TEXT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor Question Attempts: O of 3 used SAVE FOR LATER SIRAMFAKSNME

Explanation / Answer

Using force balance in vertical direction on the block

N - W - Fy = 0

N = Normal force

W = mg

Fy = F*sin A (Fy will be downward because given that F is directed BELOW the horizontal)

N = mg + F*sin A

Now Force balance in horizontal direction

Ff - Fx = 0

Fx = F*cos A

Ff = uk*N = uk*(mg + F*sin A)

F*cos A = uk*(mg + F*sin A)

F = uk*m*g/(cos A - uk*sin A)

Now using given values:

F = 0.400*38*9.81/(cos 21 deg - 0.400*sin 21 deg)

F = 188.69 N

Work-done by worker's force will be

Ww = Fx*d = 188.69*(cos 21 deg)*10 = 1761.57 J

Part B

Using Work-energy theorem:

E = Ff*d = uk*(mg + F*sin A)*d

E = 0.4*(38*9.81 + 188.69*sin 21 deg)*10

E = 1761.60 J

Please Upvote.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.