Solve Yr 11 Physics HW: Find Work Done & Energy Expended

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Homework Help Overview

The discussion revolves around a physics problem involving a car traveling up a slope, where participants are tasked with calculating the work done and energy expended. The subject area includes concepts of forces, friction, and work-energy principles.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the calculation of the car's weight, frictional resistance, and the component of weight along the slope. Questions are raised about the correct application of formulas and the interpretation of forces involved in the work done.

Discussion Status

There is an ongoing exchange of calculations and clarifications regarding the components of forces acting on the car. Some participants express confusion over their calculations, while others provide insights into the correct approach to finding the total force needed for the work done calculation.

Contextual Notes

Participants mention the use of specific formulas and the importance of SI units. There is a noted struggle with achieving correct numerical results, indicating potential misunderstandings in the application of physics concepts.

pat666
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Hey i need some help with this question please:

A car of mass 2.135 * 10^3 travels at a constant speed up a slope of 30 degrees a distance of 25.3m. If the frictional resistance is 0.100 of the weight of the car, find:
(a) the work done
(b) the energy expended

The answer is 3.18 * 10^5 J

The formulas we have been using are w=fs and f=ma also si units please.
 
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What is the weight of the car?
What is the frictional resistance?
what is the component of the weight of the car along the slope?
What is the total force?
 
i can figure the weight out using f=ma but i keep getiing the wrong answer no matter what i do??
 
here' s what I've done so far.
weight = 2.135*10^4 N
frictional resistance = 2.094*10^3
and i try to put this into w=fs cos theata
and i get 4.589*10^5

please help
 
Weight = 2.135*10^3*9.8 = 20923 N
The component of the weight of the car along the slope = 20923 N*sin(30 degree)
Total force = 20923 N*sin(30 degree) + 20923*0.100 N
Now find the work done.
 
hey thanks a lot for the help what i missed was that when you calculate the workdone it is the total force and not just the weight. thanks again
 

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