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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.

Respuesta :

ardni313
  • 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

[tex]\rm |F|^2=4^2+-3^2\\\\F=\sqrt{25}\\\\F=5[/tex]

[tex]\rm |s|=\sqrt{1^2+1^2}\\\\s=\sqrt{2}[/tex]

so that

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

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onyebuchinnaji

The work done by the force on the particle is 1 J.

The angle between the force and the displacement is 81.87⁰.

The given parameters;

  • F = (4i  - 3j)
  • displacement of the particle, s = (i + j)

The work done by the force on the particle is calculated as follows;

[tex]W = F. s\\\\W = (4i-3j) .(i + j)\\\\W = 4i^2 - 3j^2\\\\W = 4 - 3\\\\W = 1 \ J[/tex]

The angle between the force and the displacement is calculated as follows;

[tex]W =| F||s | \ cos\theta\\\\W = (\sqrt{4^2 + (-3)^2} )(\sqrt{1^2 + 1^2} ) cos(\theta)\\\\1 = 5\sqrt{2} \ cos (\theta)\\\\ cos (\theta) = \frac{1}{5\sqrt{2} } \\\\ cos (\theta) = 0.1414\\\\\theta = cos^{-1} (0.1414)\\\\\theta = 81.87^0[/tex]

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