Aforce 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.

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;

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

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

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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 | = 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

so that

velocity position

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identify the components of a vector

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mathematical statement matches the vector operation

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b.

explanation:

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