Work problem with vector notation Help please

In summary, we are given a force of 6 N in the x-direction and -3 N in the y-direction acting on a particle that undergoes a displacement of 5 m in the x-direction and 1 m in the y-direction. The work done by the force on the particle is 33.5 J and the angle between the force and displacement vectors is 348.69 degrees. Additionally, only the task of picking up a piece of chalk from the floor does not involve work out of the given options.
  • #1
od943
5
0

Homework Statement


PART A: A force ~F = Fx ˆı+Fy ˆ acts on a particle that
undergoes a displacement of ~s = sx ˆı + sy ˆ
where Fx = 6 N, Fy = −3 N, sx = 5 m, and
sy = 1 m.
Find the work done by the force on the
particle.
Answer in units of J.

PART B: Find the angle between ~F and ~s.
Answer in units of ◦.

PART C: An easy question but still need help:
Which of the following does not involve work?
1. A golf ball is struck.
2. A weight lifter does military presses (lift-
ing weights over his head.)
3. A professor picks up a piece of chalk from
the floor.
4. A runner stretches by pushing against a
wall.
5. A child is pushed on a swing.

Homework Equations


a^2+b^2=c^2
F=ma
W=fcosthetad

The Attempt at a Solution


 
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  • #2
so here's my try: i made two triangles...one out of the force vectors and another out of the displacement vectors
i found the resultant displacement vector to be 5^2+1^2=26...so 5.1 m
i found the resultant force vector by doing 3^2+6^2=45...so 6.7 N

if i plug in these numbers into the work equation of W=fdcostheta i'll get W=5.1m*6.7Ncostheta
to find theta i'll do the tantheta=1m/5m or 11.3099 degrees or 348.69 degrees when put in the correct coordinate plane...
so does W=5.1m*6.7Ncos348.69? that equals 33.5 joules.
 
  • #3

PART A: To find the work done by the force on the particle, we can use the equation W = F*s, where F is the force and s is the displacement. In this case, F = Fx ˆı+Fy ˆ and s = sx ˆı + sy ˆ. Plugging in the given values, we get W = (6 N * 5 m) + (-3 N * 1 m) = 27 J. Therefore, the work done by the force on the particle is 27 J.

PART B: To find the angle between ~F and ~s, we can use the dot product formula: cosθ = (~F * ~s) / (|~F| * |~s|), where θ is the angle between the two vectors. Plugging in the given values, we get cosθ = ((6 N * 5 m) + (-3 N * 1 m)) / (√(6 N^2 + (-3 N)^2) * √(5 m^2 + 1 m^2)) = 0.9899. Therefore, θ = cos^-1(0.9899) ≈ 8.13°. Therefore, the angle between ~F and ~s is approximately 8.13°.

PART C: Option 1, 2, and 5 all involve work, as a force is applied to an object and it undergoes a displacement. Option 3 does not involve work, as the professor is not applying a force on the chalk to make it move. Option 4 involves work, as the runner is applying a force on the wall and the wall is exerting an equal and opposite force on the runner. Therefore, the correct answer is option 3.
 

1. What is vector notation?

Vector notation is a mathematical representation of a vector using numbers and symbols. It is often used in physics and engineering to describe the magnitude and direction of a quantity.

2. How do I solve work problems using vector notation?

To solve work problems with vector notation, first identify the variables involved (such as force, displacement, and angle). Then, use the appropriate vector formulas to calculate the work done, taking into account the direction and magnitude of each vector. Finally, plug in the values and solve for the unknown variable.

3. What is the difference between scalar and vector quantities?

Scalar quantities only have magnitude (size) and no direction, such as speed or temperature. Vector quantities have both magnitude and direction, such as force or velocity.

4. Can vector notation be used to solve 3-dimensional work problems?

Yes, vector notation can be used to solve work problems in 3-dimensional space. In this case, the vectors will have 3 components (x, y, and z) instead of just 2.

5. How can I check my work when solving problems with vector notation?

You can check your work by making sure that the units of your answer match the units of the quantity you are solving for (such as Joules for work or Newtons for force). You can also use the Pythagorean theorem to check if the magnitude of your vector is correct. Additionally, double-check your calculations and make sure you have correctly accounted for direction in your answer.

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