Solving Work and Force of Wagon and Child

  • Thread starter Thread starter physicsballer2
  • Start date Start date
  • Tags Tags
    Force Work
AI Thread Summary
A child pulls a wagon with a force of 100N at a 60-degree angle, moving it 25 meters. The work done in the direction of the wagon's movement is calculated as 1250 Joules, using the formula W = FDCosθ. For part A, the force in the direction of motion must be determined, which is 50N, derived from the component of the pulling force. The confusion arises between the units of force (Newtons) and work (Joules), leading to misinterpretation of the calculations. Understanding the distinction between force and work is crucial for solving such physics problems correctly.
physicsballer2
Messages
12
Reaction score
0
A child pulls a wagon with a force of 100N on a rope that makes a sixty degree angle with a horizontal floor. He pulls the wagon a distance of 25 meters along the floor. A: how much force is done in the direction that the wagon moves? B: How much work does the child do?


W = FD
W = FDCosθ



A: W = FD

= (100)(25)
= 2500 Joules done in the direction that the wagon moves

B: W = FDCosθ

= (100)(25)Cos60
= 1250 Joules

Did I setup and solve these problems correctly?
 
Physics news on Phys.org
B is right, but A is not. A is looking for a force but you found work.
 
Is this the force that you are supposed to use for B or are you supposed to use the force that you are supposed to solve for?
 
physicsballer2 said:
Is this the force that you are supposed to use for B

Yes

or are you supposed to use the force that you are supposed to solve for?

??
 
Is there a separate equation I need to solve for the force in part A ? I assume the force they are asking for isn't given in the information. Referring to my question above should the 100 N be used for part B or should for force found in part A be used?
 
physicsballer2 said:
Is there a separate equation I need to solve for the force in part A ? I assume the force they are asking for isn't given in the information. Referring to my question above should the 100 N be used for part B or should for force found in part A be used?
Only force in direction of displacement contributes work. The force applied is at an angle to the horizontal so for part A, you need to find the component of this force in the direction of motion.
You computed B correctly, which means you effectively already have the answer.
 
I thought I understood your answer but I do not think I did. I used W=FD 1250=F(25)
F=50N

That does not seem correct
 
Think of your formula for work as W = (Fcosθ)*D

What does Fcosθ represent?
 
The force in the direction it is being pulled? Which is equal to 50 Joules
 
  • #10
physicsballer2 said:
The force in the direction it is being pulled?
Right. Which is what part A asks for.

Which is equal to 50 Joules
Newtons, not Joules. (Newton the unit of force; Joule the unit of energy.)
 
  • #11
I guess I solved wrong but got the correct answer? Because I did W = F * Cos60

W = 100 * Cos60
W = 50 joules

I appreciate your patience
 
  • #12
physicsballer2 said:
I guess I solved wrong but got the correct answer? Because I did W = F * Cos60
W = F*Cos60*25, not F*Cos60. (Work has units of force*distance, which are Joules; Force has units of Newtons.)
 
Back
Top