Two forces, vector F 1 = (-7 i + 3 j) N and vector F 2 = (-2 i + 3 j) N, act on

Question

Two forces, vector F 1 = (-7 i + 3 j) N and vector F 2 = (-2 i + 3 j) N, act on a particle of mass 2.40 kg that is initially at rest at coordinates (+1.75 m, -4.35 m). (a) What are the components of the particle's velocity at t = 11.6 s? vector v = m/s

(b) In what direction is the particle moving at t = 11.6 s? ° ( counterclockwise from the positive x axis)

(c) What displacement does the particle undergo during the first 11.6 s? vector r = m

(d) What are the coordinates of the particle at t = 11.6 s? x = m y = m

Explanation / Answer

Net force,
F = F1 + F2
= (-7 i + 3 j) + (-2 i + 3 j)
= (-9 i + 6 j) N

a = F/m
= (-9 i + 6 j)/2.4
= (-3.75i +2.5 j) m/s^2

do=(1.75 i -4.35 j) m
a)
use:
Vf= Vi + a*t
= 0 + (-3.75i +2.5 j) *11.6
= (-43.5 i + 29 j) m/s
b)
Vf=(-43.5 i + 29 j) m/s
It is 2nd quadrant.
angle = 90 + atan (43.5/29)
=146.3 degree

c)
displacement = d-do = 0.5*a*t^2
= 0.5* (-3.75i +2.5 j) *11.6^2
= (-252.3 i + 168.2 j) m
d)
use:
d = do + Vi*t + 0.5*a*t^2
d =(1.75 i -4.35 j) + 0 + 0.5* (-3.75i +2.5 j) *11.6^2
= (1.75 i -4.35 j) + (-252.3 i + 168.2 j)
= (-250.5i + 163.85 j) m
Answer: (-250.5 m ,163.85 m)

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