Question

@Abhisar
A car of mass 1390kg accelerates on a flat highway from 12 m/s to 17.0 m/s. How much work does the car's engine do on the car?

Answer 1

\(\huge \sf Work~done = Change~in~Kinetic~energy\)

Answer 2

Your response came out weird. It can't be read

Answer 3

Work done=Change in Kinetic energy

Answer 4

Right work is force times displacement . So force would be 1390xa and there's no displacements

Answer 5

If I do kinetic energy which velocity do I use

Answer 6

but you don't know a !
So u can not use W=F\(\times\)Displacement here !

Answer 7

U have to calculate change in K.E

\(\large\sf \frac{1}{2}mv^2-\frac{1}{2}mu^2\)

Answer 8

Whats (\large\sf \frac{1}{2}mv^2-\frac{1}{2}mu^2\) suppose to be

Answer 9

It just shows up as symbols on my computer

Answer 10

O_o

Answer 11

http://prntscr.com/44lhyt

Answer 12

I know kinetic is .5*139*v^2. What velocity do I use

Answer 13

U saw the screenshot ?
u r not getting u'll have to use both velocities. You have to calculate change in kinetic energy i.e final kinetic energy -initial kinetic energy.

Answer 14

Oh okay use the equation twice with each and subtract. The answer is the difference thanks

Answer 15

or u can simple plug in the values in the following equation
1/2*m*(V^2-U^2)

Answer 16

I don't know u

Answer 17

I got 101 kJ

Answer 18

u mean u don't understand me..ryt ?

Answer 19

I think I got it 101kj. Figured out what u was

Answer 20

That's correct !

Answer 21

Thanks putting up another question

Answer 22

\(\color{red}{\huge\bigstar}\huge\text{You're Most Welcome! }\color{red}\bigstar\)
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