Two forces, vector F 1 = (5.15i + 5.25j) N and vector F 2 = (2.40i + 8.10j) N, a

Question

Two forces, vector F 1 = (5.15i + 5.25j) N and vector F 2 = (2.40i + 8.10j) N, act on a particle of mass 1.80 kg that is initially at rest at coordinates (+2.05 m, 4.25 m). (a) What are the components of the particle's velocity at t = 10.2 s? vector v = m/s (b) In what direction is the particle moving at t = 10.2 s? ° counterclockwise from the +x-axis (c) What displacement does the particle undergo during the first 10.2 s? vector r = m (d) What are the coordinates of the particle at t = 10.2 s? x = m y = m

Explanation / Answer

Fnet = F1 + F2

F = 7.55i + 13.35 j

from newtons second law

Fnet = m*a

a = Fnet/m

a = (7.55/1.8)i + (13.35/1.8)j

a = 4.194 i + 7.417 j

v = vo + a*t

v = 0 + (4.194 i + 7.417 j )*10.2

v = 42.78 i + 75.65 j m/s

(b)

direction = tan^-1(75.65/42.78) = 60.5

(c)

ro = 2.05i - 4.25 j

r = ro*t + 0.5*a*t^2

r = (2.05i - 4.25 j)*10.2 + (0.5*(4.194 i + 7.417 j)*10.2^2)

r = 239.1 i + 342.5 j m

displacement = sqrt(239.1^2+342.5^2)

displacement = 417.7 m

x = 239.1 m

y = 342.5 m

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