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Why is -1/5g is nigative here

Why is -1/5g is nigative here

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Why is -1/5g is nigative here 

Wa = Fa-d = mgd cos 180° = -mg.d M Then we conclude that: * the work done by F therefore can be negative, positive or zero if the motion occurred horizontally. { 90 } Chapter 7: Kinetic Energy and Work ample V An elevator with mass of m = 500 kg is descending (moving downward) with speed vi = 4.0 m/s when its supporting cable begins to slip, allowing it to fall down with constant acceleration of a = g/5 as shown in the figure. (a) During the falling through a distance d = 12 m, what is the work Wg done on the elevator by the gravitational force F , ? (b) ) During the 12 m fall, what is the work Wr done on the elevator by the upward pull (T) of the elevator cable? (c) What is the net-work W done on the cab during the fall? The net-work is the sum ( E )of the works done by the forces acting on the cab: (d) What is the cab's kinetic energy at the end of the 12 m fall? O Solution: -Elevator cable a. The falling and gravitational force F are in the same Does direction ( 0 0), so 7 negative Cab work Wg = Fg.d = mgd cos 0° Does positive work Wg = (500 kg) (9.8) (12 m) = 58800 J × 59 kJ b. EF = ma → Fr – Fg = ma FT = ma + Fg %3D та + mg 3 т(а + g) Wt = Frd cos 0 = m(a + g)d cos 0 Does 7 negative W, = m(-8+8)d 1 g +g)d cos 0 = m g)d cos 0 Cab- work Does (4 Wr (500 kg) | × 9.8 m/2)(12 m) cos 180° = -47 kJ positive work c. W = Wg + Wr = 58800 J – 47040 J = d. W + K. E.j = K.E.f 11760 J z 12 kJ 1 2 1 (500 kg) (4- m K.E.f = ¬mv;² + W = + 11760 J K.E.f = 15760 J × 16 kJ 91 Chapter 7: Kinetic Energy and Work ork done by a spring force 1. The spring force is an example of variable force not constant, its change with the displacement. Block attached to spring x-0 F,= 0 2. When the spring at equilibrium, that means x = 0, and it's at rest , there is no compressed nor extended.
Transcribed Image Text:Wa = Fa-d = mgd cos 180° = -mg.d M Then we conclude that: * the work done by F therefore can be negative, positive or zero if the motion occurred horizontally. { 90 } Chapter 7: Kinetic Energy and Work ample V An elevator with mass of m = 500 kg is descending (moving downward) with speed vi = 4.0 m/s when its supporting cable begins to slip, allowing it to fall down with constant acceleration of a = g/5 as shown in the figure. (a) During the falling through a distance d = 12 m, what is the work Wg done on the elevator by the gravitational force F , ? (b) ) During the 12 m fall, what is the work Wr done on the elevator by the upward pull (T) of the elevator cable? (c) What is the net-work W done on the cab during the fall? The net-work is the sum ( E )of the works done by the forces acting on the cab: (d) What is the cab's kinetic energy at the end of the 12 m fall? O Solution: -Elevator cable a. The falling and gravitational force F are in the same Does direction ( 0 0), so 7 negative Cab work Wg = Fg.d = mgd cos 0° Does positive work Wg = (500 kg) (9.8) (12 m) = 58800 J × 59 kJ b. EF = ma → Fr – Fg = ma FT = ma + Fg %3D та + mg 3 т(а + g) Wt = Frd cos 0 = m(a + g)d cos 0 Does 7 negative W, = m(-8+8)d 1 g +g)d cos 0 = m g)d cos 0 Cab- work Does (4 Wr (500 kg) | × 9.8 m/2)(12 m) cos 180° = -47 kJ positive work c. W = Wg + Wr = 58800 J – 47040 J = d. W + K. E.j = K.E.f 11760 J z 12 kJ 1 2 1 (500 kg) (4- m K.E.f = ¬mv;² + W = + 11760 J K.E.f = 15760 J × 16 kJ 91 Chapter 7: Kinetic Energy and Work ork done by a spring force 1. The spring force is an example of variable force not constant, its change with the displacement. Block attached to spring x-0 F,= 0 2. When the spring at equilibrium, that means x = 0, and it's at rest , there is no compressed nor extended.
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