2:381 For this assignment you submit answers by question parts. The number of su

Question

2:381 For this assignment you submit answers by question parts. The number of submissions remaining for each question part only changes if you submiê or change the anwer Assignment Scoring Your last submission is used for your score Two forces, il . (S-5j) NandF2.(21-4j) N, act on a partide of mass 2.20 kg that is initialy at rest at coordinates (-1.60 m, +4.50 m). (a) What are the components of the particle's velocity at t- 10.2s -(32 43-41718) (b) In what direction is the particle moving at t 10.2s Your response differs from the correct answer by more than 10%. Double check your calculations.° ( counterclockwise from the positive x axis) (c) What displacement does the particle undergo during the first 10.2 s (d) What are the coordinates of the particle at t-10.2 s? A block slides down a frictionless plane having an inclination of 8-13.2 The block starts from rest at the top, and the length of the indline is 2.45 m. (a) Draw a free-body diagram of the block This answer has not been graded yet (b) Find the acceleration of the block m/s2 down the incline (c) Find its speed when it reaches the bottom of the incline. m/s The systems shown below are in equilibrium with m-2.20 kg and 8 5.0°. If the spring scales are calibrated in newtons, what do they read? Ignore the masses of the pulleys and strings and assume the pulleys and the ncline are frictionless.

Explanation / Answer

1. given

F1 = 5i - 5j N

F2 = 2i - 4j N

m = 2.2 kg

ro = (-1.6, 4.5) m

a. at t= 10 .2 s

v = u + at

v = (F1 + F2)*t/m = (7i - 9j)*10.2/2.2 = 32.4545i - 41.72727j m/s

b. at t = 10.2 s

tan(theta) = (-41.727/32.454)

theta = -52.12530 deg

hence the particel is travelling 52.1250 deg below +x axis

c. r = ro + ut + 0.5at^2

r = -1.6i + 4.5j + 0.5(7i - 9j)*(10.2)^2/2.2

r = 163.91818i -208.30909j m

disp = r - ro = r + 1.6i - 4.5j = 165.51818i -212.80909j

|disp| = 269.59966004 m

d. coorindates = (163.918, -208.309) m

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