Change in the kinetic energy of a car driving up a hill

AI Thread Summary
The discussion focuses on calculating the change in kinetic energy of a car driving up a hill, emphasizing that the value of velocity (v) is not needed for this specific calculation. Participants clarify that the resistive force of 440 is used to determine the work done against it, which is essential for calculating the total energy required to move the car uphill. The correct approach involves considering both gravitational potential energy and work against resistive forces. The final calculation yields a time of approximately 3.5 seconds for the car to ascend the hill, confirming the accuracy of the computations. Overall, the thread highlights the importance of understanding forces and energy in motion for solving physics problems.
hello478
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Homework Statement
mentions below in picture
Relevant Equations
energy equations
and work done = f*d moved in direction of force
part i)
i did 1/2 * 1700 * v^2
i dont know what v is...
so how do i solve it?

part ii)
i calculated it correctly by 440*25
please explain in detail why i used 440?

and part d)
i did 1.7*10^4 = 48000/t
my t= 2.82 s
but correct answer is 3.5s

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hello478 said:
i dont know what v is...
You aren't being asked to calculate the kinetic energy, you are being asked how much it changes. You don't need the value of ##v## to perform that calculation.
please explain in detail why i used 440?
You are being asked to calculate the work done against the resistive force. What is that resistive force?
and part d)...
The work being done against the resistive force is not the only work that has to be done to get the car up the slope. Find the work that you haven't counted, include it, see what you get.
 
so for the first one
change in KE =0

resistive force is 440 so thats why we use it

and for part d
the total energy would be
gpe + the work done against resistive force
so 48000+440*25 / 17000
so time = 3.47 which is almost equal to 3.5
please tell me if im wrong
and thank you so much for your answer
 
Seems about right.
 
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