Question

help!!! ASAP

Answer 1

A car has a total energy of 120000 J as it drives up a hill. If the car has a mass of 1200kg and is traveling at 2m/s, how high is it up in the hill? Define position A and Position B?

Answer 2

@ganeshie8 help please!

Answer 3

@AakashSudhakar please help lol

Answer 4

also can you help me drawing it into a bar graph for position a and position b

Answer 5

@AakashSudhakar are you almost done typying? sorry

Answer 6

@AakashSudhakar are you almost done?

Answer 7

This is related to conservation of energy. Recall that energy is always theoretically conserved in a system. This means that the conservation of energy is given by... \[PE _{hilltop} = PE _{hill} + KE _{hill}\]The potential energy of the hilltop is the conserved total energy, which we are given as 120000 J. [As an engineering major, I'm going to take the liberty of denoting that quantity from this point forth as 1.2 x 10^5 J.] This means that... \[PE _{hill} + KE _{hill} = (1.2\times10^{5})Joules\]We can also recall the definitional equations of physical potential energy and physical kinetic energy: \[PE = mgh\]\[KE = \frac{ 1 }{ 2 }mv ^{2}\]Therefore, we can finally make the statement that... \[mgh + \frac{ 1 }{ 2 }mv ^{2} = (1.2\times10 ^{5}) Joules\]You're give that the car's mass is m=1200kg, the car's velocity is v=2m/s, and you know that the gravitational acceleration on Earth is g=9.8m/s^2. Therefore, solving for h gets you the solving equation:\[h = \frac{ (1.2\times10 ^{5} Joules) - \frac{ 1 }{ 2 }mv ^{2} }{ mg }\]Plug in everything you have and solve for h. Unfortunately, I can't do the bar graph for you, but you have enough information to get that.

Answer 8

In the frame of reference that I solved the problem in, Position A would be the current height of the car and Position B would be the height of the top of the hill.

Answer 9

for the height i got 9.98 is that correct @AakashSudhakar

Answer 10

That should be correct. I obtained around 10 meters as a result, which is an approximation from 9.98 meters.

Answer 11

can you please help me with the bar graph? i need help

Answer 12

@AakashSudhakar