Work and Power: Lever Efficiency Problem

  • Thread starter Thread starter NDiggity
  • Start date Start date
  • Tags Tags
    Power Work
AI Thread Summary
A physics problem involves pulling a 1000kg boat 50m with a 40N force at a 45-degree angle. To calculate work done, the horizontal component of the force must be considered, leading to the formula W = F.d cos(θ). The correct calculation for work is 40N * 50m * cos(45°), not simply 40N * 50m. Additionally, a question about the efficiency of a lever system reveals confusion in calculating the distance moved by the effort force, with the user isolating variables incorrectly, resulting in an implausibly small distance. Understanding the role of angles and efficiency in work calculations is crucial for accurate problem-solving.
NDiggity
Messages
53
Reaction score
0

Homework Statement


A person uses a rope to pull a 1000kg. boat 50.0m along a wharf. The rope makes an angle of 45 degrees with the horizontal. If a force of 40.0N is used to move the boat, how much work is done?

Homework Equations


W=fd
Here is my problem. My physics class has 5 days left and we have 3 units to get through so were basically doing a unit in a day and a half. So my teacher is giving a very brief overview of the units and then gives us work sheets. So I am still a little confused. Is the question just giving me extra information to trick me? My question is, if the direction of movement is horizontal like in this question, do you use the horizontal force to find work then? Do you only use the tension in the rope if it is given to find the force needed to move the box horizontally.

The Attempt at a Solution


I'm not too sure. I thought maybe it was 40(50)=2000J since 40N is needed to move the boat 50m.
 
Last edited:
Physics news on Phys.org
You are close but not close enough. Remember that the rope makes an angle with the horizontal.
 
is it 40(50)cos45? I'm obviously still a little confused with this unit since we had to learn the whole thing in a day.
 
NDiggity said:
is it 40(50)cos45?
That is correct.

The work done by a constant force is F.d, the dot product of the force and displacement. In the problem at hand, the displacement is purely horizontal, while the force has both horizontal and vertical components.

Here's a site that you might find helpful, http://hyperphysics.phy-astr.gsu.edu/hbase/wcon.html
 
correct:biggrin:
 
Thank you very much for the help! I have one more question.

A force of 2250N is exerted on a lever system to raise a 1200N rock 1.32m. If the efficiency of the lever is 58.7%, how far did the effort force move?

So I used the efficiency equation: Efficiency=work with no machine/work with machine

I plugged in the variables to get:
58.7=1200(1.32)/2250d

I isolated d and got .012m. This seems too small. What did I do wrong?
 
Last edited:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top