Question

A car and truck have kinetic energy of 8×10^5J and 9×10^5J respectively. lf they are brought to rest after covering the same distance, find the ratio of forces applied to both vehicles​

Answer 1

hey!!!

Mark it as brilliant if it is usefull

Let MM and VV be respectively the mass and velocity of the truck and mm and vv be the mass and velocity of the car. Also, let us assume that they both have wheels made of the same material, which has a friction coefficient μμ with the asphalt. If both of them have the same kinetic energy, we can say for sure that

T=12MV2=12mv2T=12MV2=12mv2

And, consequently,

v2V2=Mm (1)v2V2=Mm (1)

Now, we know that there’s only one force acting on each body on the horizontal direction — since the weight and the floor’s normal reaction cancel each other. That would be the friction force, which, for the car, is given by

F=μN=μmg (2)F=μN=μmg (2)

We should note that the work done by friction is the one responsible for converting the car’s kinetic energy to other forms of energy — which in this case should be heat. As such, we can say that

W=ΔT⇒F⋅d=12mv2W=ΔT⇒F⋅d=12mv2, where d is the distance traveled by the car.

Plugging in (2) into that, we get

μmgd=12mv2μmgd=12mv2

⇒d=v22g⇒d=v22g

In an analogous manner we find the distance DD traveled by the truck:

D=V22gD=V22g

Dividing both equations, we have that

dD=v2V2dD=v2V2

However, by (1), we can state that

dD=MmdD=Mm

Which is the general solution to your problem. By common sense, we should have the truck to be heavier than the car, which implies that M>mM>m and therefore Mm>1Mm>1. That in turn would assure that dD>1dD>1, which is to say, d>Dd>D. Thus, we prove that the distance traveled by the truck should be shorter than the distance traveled by the car, if the previous is heavier than the latter.

Answer 2

Hiiiiiiiiiiiii

Mateeeeeeeeeee

.....