VA Chapter 5 HW C O www.webassign. t/web/Student/Assignme Response /submit? dep

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VA Chapter 5 HW C O www.webassign. t/web/Student/Assignme Response /submit? dep 5409492 12. O 016 points I Previous Answers HRW7 5.P 026 My Notes Figure 5-40 shows an overhead view of a lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force F1 has a magnitude of 6.00 N and is at 61 24 Force F2 has a magnitude of 7.00 N and is at 62 24°. The lemon half has mass 0.0200 kg Fig. 5-40 What is the third force if the lemon half has the following velocities? (a) zero X j N (b) constant velocity v 13.0 i 14.0 j) m/s (c) varying velocity v 13,0t i 14.0t j) m/s, where t is time in seconds X j) N Section 5.6 Newton's Second Law Section 2.6 Acceleration Submit Answer Save Progress 13- -13 points HRW7 5.P029 My Notes A block is projected up a frictionless inclined plane with initial speed Vo 3 3.49 m/s. The angle of incline is e 3 32.2°. a A 4) Search here 2/13/2017 R6

Explanation / Answer

here,

force 1, f1x = 6 * cos24
force 1, f1y = 6 * sin24
force, f1 = (6*cos24 i + 6*sin24j)

force 2, f2x = 7 * cos24
force 2, f2y = 7 * sin24
force, f2 = (7*cos24 i + 7*sin24j)

Part a:
since instantaneous velocity is zero, so third force will be,

f1 + f2 + f3 = 0

(6 * cos24 + 7 * cos24)i + (7 * sin24 + 6 * sin24)j + f3 = 0

f3 = (-11.876 i - 5.288 j)

part b:
Velocity is constant. Therefore F3 should be such that net force = 0.

so, f3 = (-11.876 i - 5.288 j)

part c:
velocity, v = (13t i - 14t j)

acceleration, a = (13i - 14j)

f1 + f2 + f3 = mass * acceleration

(6 * cos24 + 7 * cos24)i + (7 * sin24 + 6 * sin24)j = (13i - 14j)

(11.876 i + 5.288 j) + f3 = 0.02 * (13i - 14j)

F3 = (-11.616 i) - (5.568) j

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