How much work does the brake system have to do to stop the car at each speed?

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Homework Help Overview

The discussion revolves around calculating the amount of work required by a car's brake system to stop the vehicle at various speeds, specifically 60 km/h, 50 km/h, 40 km/h, 30 km/h, 20 km/h, and 10 km/h. The original poster has calculated the kinetic energy for a car with a mass of 1500 kg at these speeds but is uncertain about how to determine the work done by the brakes.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between kinetic energy and work done by the brakes, with some questioning the units used in the calculations. The original poster attempts to clarify their understanding of kinetic energy and its units.

Discussion Status

The conversation is ongoing, with participants raising questions about unit consistency and the definitions of kinetic energy. There is no explicit consensus reached, but there is a focus on ensuring correct unit usage and understanding the underlying physics concepts.

Contextual Notes

Participants are addressing potential misunderstandings related to the units of measurement for kinetic energy and the work-energy principle. The original poster is working within the constraints of a homework assignment that requires specific calculations without provided equations for the second part of the question.

laylay918
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Homework Statement



Investigate the amount of kinetic energy involved when your car's speed is 60 km/h, 50 km/h, 40 km/h, 30 km/h, 20 km/h, and 10 km/h. (Find your car's mass in the owner's manual) How much work does the brake system have to do to stop the car at each speed?

Homework Equations


KE = 1/2 mv^2

The Attempt at a Solution



I figured out the KE for each of those speeds for a 1500 kg car. The problem is the second portion of the question. How much work does the brake system have to do to stop the car at each speed? I have no idea how to solve that part because no equations are given. Anyways, here is my work for the first portion.KE = 1/2(1500kg)(60km/h)^2 = 2,700,000 J
KE = 1/2(1500kg)(50km/h)^2 = 1,875,000 J
KE = 1/2(1500kg)(40km/h)^2 = 1,200,000 J
KE = 1/2(1500kg)(30km/h)^2 = 675,000 J
KE = 1/2(1500kg)(20km/h)^2 = 300,000 J
KE = 1/2(1500kg)(10km/h)^2 = 75,000 J
 
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At 10 km/hr, the KE is 75 kJ. When the car is at rest how much kinetic energy does it have?

Hence how much work will the braking system need to do to get the KE to this value?
 
Make sure you check your units.

You are calling a kg (km/hr)2 a Joule. What are the units for Joules?
 
Villyer said:
Make sure you check your units.

You are calling a kg (km/hr)2 a Joule. What are the units for Joules?

No, I'm calliing ".5mv^2" a Joule.
 
laylay918 said:
No, I'm calliing ".5mv^2" a Joule.
Hello laylay918. Welcome to PF !

That's kinetic energy. If the mass, m, is in units of kg and the velocity, v, is in units of meters/second, the unit for (1/2)mv2 will be a Joule.
 
laylay918 said:
No, I'm calliing ".5mv^2" a Joule.

I don't mean mathematically, I mean in units.
 

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