in a game a toy car slides down a slop if the top of the slope is 2 m higher than the foot of the slope how fast will the car be moving when it reaches the foot assume that all of its g.p.e is transformed to k.e (i'm 8th grade so simple answers pleas)

since PE=mgh and KE = 1/2 mv^2,

gh = 1/2 v^2
(9.8)(2) = 1/2 v^2
. . .

thx man

I really needed help from that. I thought there wasn't enough information but thanks a lot!

To find the speed at which the toy car will be moving when it reaches the foot of the slope, you can use the principle of conservation of energy.

The potential energy of the car at the top of the slope is transformed into kinetic energy at the foot of the slope.

Let's assume that the mass of the toy car is "m" (which we don't know), the height of the slope is "h" (which is given as 2 meters), and the speed of the car at the foot of the slope is "v" (which we need to find).

The potential energy (PE) of the car at the top of the slope can be calculated using the formula:

PE = m × g × h

Where "g" is the acceleration due to gravity (approximately 9.8 m/s^2).

At the foot of the slope, all of the potential energy is transformed into kinetic energy (KE). The kinetic energy (KE) of the car can be calculated using the formula:

KE = (1/2) × m × v^2

Since the potential energy is transformed into kinetic energy, we can equate the two:

PE = KE

m × g × h = (1/2) × m × v^2

Now, we can solve the equation to find the speed:

m × g × h = (1/2) × m × v^2

Canceling out the mass (m) on both sides of the equation:

g × h = (1/2) × v^2

Substituting the known values:

(9.8 m/s^2) × (2 m) = (1/2) × v^2

Simplifying further:

19.6 m^2/s^2 = (1/2) × v^2

To find "v", we can rearrange the equation:

v^2 = 2 × 19.6 m^2/s^2

v^2 = 39.2 m^2/s^2

Finally, taking the square root of both sides:

v = √39.2 ≈ 6.26 m/s

Therefore, the toy car will be moving at approximately 6.26 meters per second when it reaches the foot of the slope.