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Two forces, F, = (-4.00î – 3.95j) N and F, = (-3.95î - 4.95j) N, act on a particle of mass 1.90 kg that is initially at rest at 1 coordinates (+1.95 m, +4.05 m). (a) What are the components of the particle's velocity at t = 11.1 s? m/s (b) In what direction is the particle moving at t = 11.1 s? ° counterclockwise from the +x-axis (c) What displacement does the particle undergo during the first 11.1 s? AF = m (d) What are the coordinates of the particle at t = 11.1 s? y = m

Two forces, F, = (-4.00î – 3.95j) N and F, = (-3.95î - 4.95j) N, act on a particle of mass 1.90 kg that is initially at rest at 1 coordinates (+1.95 m, +4.05 m). (a) What are the components of the particle's velocity at t = 11.1 s? m/s (b) In what direction is the particle moving at t = 11.1 s? ° counterclockwise from the +x-axis (c) What displacement does the particle undergo during the first 11.1 s? AF = m (d) What are the coordinates of the particle at t = 11.1 s? y = m

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Two forces, F :(-4.00î – 3.95ĵ) N and F, = (-3.95î – 4.95j) N, act on a particle of mass 1.90 kg that is initially at rest at 1 coordinates (+1.95 m, +4.05 m). (a) What are the components of the particle's velocity at t = 11.1 s? m/s (b) In what direction is the particle moving at t = 11.1 s? ° counterclockwise from the +x-axis (c) What displacement does the particle undergo during the first 11.1 s? (d) What are the coordinates of the particle at t = 11.1 s? X = m y =
Transcribed Image Text:Two forces, F :(-4.00î – 3.95ĵ) N and F, = (-3.95î – 4.95j) N, act on a particle of mass 1.90 kg that is initially at rest at 1 coordinates (+1.95 m, +4.05 m). (a) What are the components of the particle's velocity at t = 11.1 s? m/s (b) In what direction is the particle moving at t = 11.1 s? ° counterclockwise from the +x-axis (c) What displacement does the particle undergo during the first 11.1 s? (d) What are the coordinates of the particle at t = 11.1 s? X = m y =
Expert Solution
Step 1

Solution:

Given 

F1⇀=(-4.00i ⏜-3.95j⏜)  NF2⇀=(-3.95i ⏜-4.95j⏜)  Nmass m=1.90 kgcoordinates =(+1.95 m +4.05 m)

(a) Net force on the particle is

F⇀=F1⇀+F2⇀

F⇀=(-4i⏜-3.95j⏜)+(-3.95i⏜-4.95j⏜)F⇀=(-7.95i⏜-8.9j⏜) N

we know that

acceleration= a⇀=F⇀m=(-7.95i⏜-8.9j⏜) N1.90 kg

∴a⇀=(-4.18i⏜-4.68j⏜ ) m/s2

So,

v⇀=u⇀+at⇀v⇀=0+(-4.18i⏜-4.68j⏜) m/s2 (11.1 s)v⇀=(-46.39i⏜-51.94j ⏜) m/s

x-component = -46.39 m/s

y-component = -51.94 m/s

Therefore,

The components of the particle's velocity at t=11.1 S is

v⇀=(-46.39i⏜-51.94j ⏜) m/s

 

(b)

We know that

tanθ =vyvx

tan θ=-51.94-46.39=1.11θ =tan-11.11∴θ =47.98o

Hence,

The particle is moving at an angle  θ =47.98o

counter clockwise from the +x-axis.

 

 

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