Question

A force F~ = Fx ˆı + Fy ˆj acts on a particle that undergoes a displacement of ~s = sx ˆı + sy ˆj where Fx = 4 N, Fy = −3 N, sx = 1 m, and sy = 1 m. Find the work done by the force on the particle. Answer in units of J.
Find the angle between F~ and ~s.

Answer 1

• The work done by the force on the particle : W = 1 J
• The angle between F and s : 81.870°

Further explanation

Vector is a quantity that has a value and direction

Vector can be symbolized in the form of directed line segments

The magnitude of the vector is denoted by |a|

The dot product between 2 vectors a and b can be defined:

a. b = | a || b | cos teta

| a | = length of a

| b | = length b

θ = angle between a and b

For 2 unit vectors

a = [a1, a2] ⇒ axi + ayj

b = [b1, b2] ⇒ bxi + byj

then

a. b = axi. bxi + ayj. byj

a . b = ax bx + ay by

with

| a |² = ax² + ay²

| b |² = bx² + by²

Work is a dot product between force and object movement

W = F. S.

or

W = | F || s | cos θ

From the problem we get the unit vector of the force and its displacement:

F = 4i + -3j

s = i + j

so that

W = F. s

W = (4i + -3j) (i + j)

W = 4i.i + -3j.j

W = 4.1 + -3.1

W = 1 J

Angle between F and S:

W = | F || s | cos θ

$\rm |F|^2=4^2+-3^2\\\\F=\sqrt{25}\\\\F=5$

$\rm |s|=\sqrt{1^2+1^2}\\\\s=\sqrt{2}$

so that

$\rm 1=5.\sqrt{2}\:cos\:\theta\\\\cos\:\theta=\dfrac{1}{5.\sqrt{2}}\\\\\theta=\boxed{\bold{81,870^o}}$

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Answer 2

force in component form is given as

F~ = Fx ˆı + Fy ˆj

given that : Fx = 4 N, Fy = −3 N

hence the force equation looks like

F~ = 4 ˆı + (- 3) ˆj

F~ = 4 ˆı - 3 ˆj

displacement is given as

~s = sx ˆı + sy ˆj

given that : sx = 1 m, and sy = 1 m

hence the displacement is given as

~s = 1 ˆı + 1 ˆj

work done is the dot product of force and displacement . hence work done is given as

W = F~.~s

W = (4 ˆı - 3 ˆj) . (1 ˆı + 1 ˆj )

W = (4 x 1) ( ˆı.ˆı) - (3 x 1) (ˆj.ˆj)

W = 4 - 3

W = 1 J

θ = angle between F~ and ~s = ?

hence the work done by the force comes out to be 1 J

|F~| = magnitude of force = sqrt(Fx² + Fy²) = sqrt((4)² + (-3)²) = 5 N

|~s| = magnitude of displacement = sqrt(sx² + sy²) = sqrt((1)² + (1)²) = sqrt(2)

we know that work done is given as

W = |F~| |~s| Cosθ

1 = (5) (sqrt(2) Cosθ

Cosθ = 0.142

θ = Cos⁻¹(0.142)

θ = 81.84 deg