Total work of a directional wind on a mailman

[ac7597]

Homework Statement

Mike the Mailman takes his oath seriously: "Neither snow, nor rain, nor heat, nor gloom of night stays these courageous couriers from the swift completion of their appointed rounds". Even though a blizzard is raging outside, he goes out to deliver the mail.

He makes four stages along his route:

First, he walks 40 meters North.

Next, he walks 53 meters East.

Then, he walks 42 meters at an angle of 30 degrees South of East.

Finally, he walks 80 meters at an angle of 10 degrees West of South.

The entire time he is outside, the wind pushes him with a force of 130 Newtons at at 36 degrees South of East,

How much work does the wind do to Mike over the course of his deliveries?

What is Mike's displacement from his original position? Express your answer in terms of vector components:

Relevant Equations

net work= force(distance)cos(theta)

Homework Statement: Mike the Mailman takes his oath seriously: "Neither snow, nor rain, nor heat, nor gloom of night stays these courageous couriers from the swift completion of their appointed rounds". Even though a blizzard is raging outside, he goes out to deliver the mail.

He makes four stages along his route:

First, he walks 40 meters North.

Next, he walks 53 meters East.

Then, he walks 42 meters at an angle of 30 degrees South of East.

Finally, he walks 80 meters at an angle of 10 degrees West of South.

The entire time he is outside, the wind pushes him with a force of 130 Newtons at at 36 degrees South of East,

How much work does the wind do to Mike over the course of his deliveries?

What is Mike's displacement from his original position? Express your answer in terms of vector components:
Homework Equations: net work= force(distance)cos(theta)

I initially tried to substitute distance=40+53+42+80. Force=130N and theta=36degree. The total work=(130N)(215m)cos(36)=22.6 * (10^3) J.

 

[tnich]

Suppose the mailman adds one more leg to his trip and walks the shortest route back to his starting point. Would you expect the total work done by the wind on the mailman to be zero at that point?

 

[ac7597]

distance is the displacement from the starting point to end point. I got the vector components as 75.5i-59.8j. The displacement is (75.5^2 + 59.8^2)^(1/2)=96.285. This gives net work=130N(96.285m)cos(36)=10.1E3 J.

 

[ac7597]

10.1E3 J is wrong. I don't know why.

 

[Delta2]

10.1E3 J is wrong. I don't know why.

Assuming that the total displacement vector is $75.5i-59.8j$ (haven't checked that sorry), you need to find the angle $\phi$ that this vector makes with the vector of wind. Then the work will be $130\sqrt{75.5^2+59.8^2}\cos\phi$.

 

[ac7597]

(105.17i-76.4j) is the vector component of 130N. Thus (75.5i−59.8j)*(105.17i-76.4j)=|130||96.285|cosϕ.
ϕ=2.17deg. Thus 130N(96.285m)cos(2.17)=12.5E3 J. This is correct thanks.