Two forces, vector F 1 = (5.70i + 4.60j) N and vector F 2 = (3.45i + 8.10j) N, a
ID: 1593633 • Letter: T
Question
Two forces, vector F 1 = (5.70i + 4.60j) N and vector F 2 = (3.45i + 8.10j) N, act on a particle of mass 2.50 kg that is initially at rest at coordinates (1.50 m, +3.65 m). (a) What are the components of the particle's velocity at t = 10.6 s? vector v = 38.80 i +61.48 j Incorrect: Your answer is incorrect. m/s (b) In what direction is the particle moving at t = 10.6 s? Incorrect: Your answer is incorrect. ° counterclockwise from the +x-axis (c) What displacement does the particle undergo during the first 10.6 s? vector r = 205.62 i +325.84 j Incorrect: Your answer is incorrect. m (d) What are the coordinates of the particle at t = 10.6 s? x = m y = m Need Help? Read It
Explanation / Answer
(a) F 1 = (5.70i + 4.60j) N
F 2 = (3.45i + 8.10j) N
Ff = F1 + F2
Ff = (9.15i + 12.70j)
F = ma;
9.15 = 2.5ai
ai =3.66 m/s^2
F = ma;
12.70 = 2.5a
aj = 5.08 m/s^2
V = 3.66(10.6)i + 5.08(10.6)j
V = 38.796i + 53.848j m/s
(b) In what direction is the particle moving at t = 10.6s
Vector F = sqrt ( (9.15)^2 + 12.7^2) = root(245)
tetha = arctan (12.7/9.15)
theta = 54.23 deg...........Ans.
(c) what displacement does the particle undergo during the first 10.6s
Xf = Xi + VixT + 1/2AxT^2
deltaX = 0+ 1/2(3.66)(10.6)^2
deltaX = 205.62 m ........Ans.
Xf = 205.62 -1.5 = 204.12 m
Yf = Yi + ViyT + 1/2AyT^2
delta Y = 0 + 1/2(5.08)(10.6)^2 = 285.40 m ...........Ans.
Yf = 285.40 + 3.65 = 289.05 m.
(204.12i, 285.40j)
(d) (204.12i, 285.40j)...........Ans.
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