Work done using time, velocity and mass?

In summary, the law of conservation of energy applies to the motion of vehicles. The work done on a 1200kg vehicle when it slows from 90kmh-1 to 50kmh-1 in 8.0 seconds can be calculated using the formula W=1/2mv^2 - 1/2mu^2, resulting in an answer of -2.59×105 Joules. However, it is important to convert the speeds given in kmh-1 to ms-1 before plugging them into the formula.
  • #1
TDizzl
12
0

Homework Statement



The law of conservation of energy applies to the motion of vehicles. Find the work done on a 1200kg vehicle when it slows from 90kmh-1 to 50kmh-1 in 8.0 seconds.

Homework Equations



W=1/2mv^2 - 1/2mu^2

The Attempt at a Solution


[/B]
W=0.5*1200*50^2 - 0.5*1200*50^2
 
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  • #2
If you assume a constant deceleration, you can use constant acceleration equations such as x = 1/2(vo + v)*t and v = vo + a*t. Work = Fd, where F is the force and d is the displacement. This can be expanded to W = m*a*d. Using the acceleration equations, you should be able to find all of your unknowns and plug them into the work equation.
 
  • #3
TDizzl said:
W=0.5*1200*50^2 - 0.5*1200*50^2
Is that what you meant to post? What happened to the 90kmph?
cnstntcnfsn said:
If you assume a constant deceleration...
Why would TDizzl care what the rate of deceleration is?
 
  • #4
Yes, my bad
Its W=0.5*1200*50^2 - 0.5*1200*90^2
 
  • #5
cnstntcnfsn said:
Using the acceleration equations

The question is from a paper which does not require the use of constant acceleration equations. It would seem bizarre to use a formula that hasn't been introduced to us yet.
Also, the solution is written as -2.59*10^5 Joules in the paper.
 
  • #6
TDizzl said:
Yes, my bad
Its W=0.5*1200*50^2 - 0.5*1200*90^2

You're not proposing to leave that as your answer, are you?
 
  • #7
PeroK said:
You're not proposing to leave that as your answer, are you?
No, that was how I substituted the values in as an attempt to answer the question.
 
  • #8
TDizzl said:
Yes, my bad
Its W=0.5*1200*50^2 - 0.5*1200*90^2
In what units?
 
  • #9
haruspex said:
In what units?
m=1200kg
v=50kmh-1
u=90kmh-1
 
  • #10
TDizzl said:
m=1200kg
v=50kmh-1
u=90kmh-1
So what is it in Joules?
 
  • #11
haruspex said:
So what is it in Joules?
Yea its like -3360000J, that's what I got
 
  • #12
TDizzl said:
Yea its like -3360000J, that's what I got
No, the speeds you are given are in kmh-1, not ms-1.
 
  • #13
So how do I get the answer as -2.59×105?
 
  • #14
TDizzl said:
So how do I get the answer as -2.59×105?
How do you convert km/h to m/s?
 
  • #15
Alright -2.59×105 is the answer I got, the problem was that I was doing 502/3.6 instead of (50/3.6)2.
Thanks for the help, feeling extremely stupid. :mad:
 

1. How do you calculate work done?

The formula for calculating work done is: Work = Force x Distance. This means that in order to calculate the work done, you need to multiply the force applied to an object by the distance the object moves in the direction of the force.

2. What is the role of time in calculating work done?

Time is not directly involved in the calculation of work done. However, time can be used to calculate the average velocity of an object, which is then used in the formula for work done. Time also plays a role in determining the rate at which work is done, known as power.

3. How does mass affect the amount of work done?

Mass is a factor in determining the amount of work done. The greater the mass of an object, the more force is required to move it a certain distance, resulting in more work done. This is because mass is directly proportional to the force required to move an object.

4. How is work related to velocity?

Velocity is also a factor in calculating work done. Work done is directly proportional to the velocity of an object. This means that the faster an object is moving, the more work is being done on it.

5. Can work be negative?

Yes, work can be negative. This happens when the force applied to an object is in the opposite direction of the object's motion. In this case, the work done is considered to be negative because the force is actually taking away energy from the object rather than adding it.

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