Question

Two forces, vector F 1 = (−4.20î − 3.65ĵ) N and vector F 2 = (−2.30î − 7.60ĵ) N, act on a particle of mass 1.90 kg that is initially at rest at coordinates (+1.75 m, −3.95 m).

(a) What are the components of the particle's velocity at t = 11.3 s?

(b) In what direction is the particle moving at t = 11.3 s?

° counterclockwise from the +x-axis

(c) What displacement does the particle undergo during the first 11.3 s?

(d) What are the coordinates of the particle at t = 11.3 s?

x =

y =

Answer 1

a)

F1 = -4.20 i 3.65 j

F2 =  -2.30 i - 7.60 j

using the Newton second law we can find the acceleration a = f / m so m = 1.90kg

F = F1 + F2;

F = -6.6 i - 3.95 j

so the acceleration a = ( -3.474 i - 2.08 j ) m/s^2.

we know that v(t) = a*t

so the particle's velocity at t = 11.3 s is

V = ( -39.2562 i - 23.92 j ) m /s.

b)

theta = tan^-1(vj/vi) = tan^-1(23.92/39.2562) = 31.36 degree negative x direction

c)

x = 1/2at^2

x = 0.5 x ( -3.474 i - 2.08 j ) x 11.3^2

x = -221.79 i - 132.79 j

d)

the coordinates are

xf = x + xi

xf = -221.79 i - 132.79 j + 1.75 i - 3.95 j

xf = -220.04 i - 136.74 j