# A toy car slides down a slope. If the top of the slope is 2m 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?

Assuming there are no energy losses due to friction or drag, the gravitational potential energy will change into kinetic potential energy as the car reaches the bottom of the slope. G.P.E = m*g*h K.E = (m*v^2)/2 where m = mass of toy car (kg) g = gravity (m/s^2) h = heigh of your car from the bottom (m) v = velocity of the toy car as it reaches the bottom (m/s) Equate K.E to G.P.E G.P.E = K.E m*g*h = (m*v^2)/2 make v the subject of the formula v = (2*g*h)^(1/2) Substitute g = 9.81 m/s^2 and h = 2m into the equation to get v v = (2*9.81*2)^(1/2) v = 6.264 m/s

The speed of the car is 6.26m/s^2

Yes calculate slope

Conservation of energy means GPE = KE mgh = 0.5 m square(v) Thus, v = square root (2gh) where V = velocity of toy car g = acceleration due to gravity h = height of slop I m typing the answer using smartphone, so sorry for my writing format but I hope you will get the meaning and itβs usage.

This one is a great and common question. If you just want the answer without explanation skip to the bottom and you will get it. This is the theory of conservation of energy. That means that energy cannot be created or destroyed; so all of the energy in the universe was here at the beginning is the same energy that is here now. Therefore we now have to see in this question how energy was changed from one type to another or how it was βtransferredβ from one to another. There are a few key words when we are talking about energy. When we talk about βheightβ and βenergyβ together we are almost always talking about gravitational potential energy or GPE for short. The equation for this is: GPE = mass x gravitational acceleration x height GPE = mgh If we substitute the numbers into the equation and calculate the energy we get: GPE = m x 9.81 x 2 GPE = 19.62m We do not know the mass of the car but itβs okay at this point. The second part of the question indicates that we are looking for the speed or velocity AND energy. This type of energy is called kinetic energy. Any thing that moves has kinetic energy. KE = 1/2 mv^2 To show that all energy has been transferred we have to make both equations equal to each other. KE = GPE 1/2mv^2 = mgh 1/2mv^2 = 19.62m 1. Divide by m. 1/2v^2 19.62 2. Divide by 1/2. v^2 = 39.24 3. Square root both sides. v = 6.264m/s

According to conservation of mechanical energy the potential energy of toy car at top of the slope must be equal to kinetic energy of the car at foot of the slope Therefore 1/2 mv^2 = mgh V = (2gh)^1/2 By solving above equation we get V = 6.26 m/s Or we can also use the formula v^2 - u^2 = 2as But in place of 's' & a we have to substitute s = h / sin(theta) & a = g sin(theta) Finally we get the equation as v = (2gh)^1/2 Note: Where theta is the angle made by slope with base.

You can simple use SUVAT to answer this question. The acceleration is 9.8m/s^2 due to gravity and the initial velocity is 0. We are trying to calculate the final velocity and we know the distance is 2m. Simply using v^2=u^2+2as Where v is final velocity( aka final speed), u is initial velocity, a is acceleration and s is the distance. The answer will 6.26 ( 3sf)

6.3 metre per sec

This slope question is a regular energy conservation question. We must first figure out the change in gravitational potential energy (g.p.e) which is deltaE=my * delta h. We also know that all g.p.e is transformed into kinetic energy (k.e) which means deltaEk = 1/2 mv^2. We can calculate the gpe first and separately but it is always more efficient and accuarate to work with symbols and substitute at the end therefore: Gpe=ke =>> mg delta h = 1/2 mv^2 <==> sqrt(2g delta h) = v ( note take the positive root as they want speed) Now we sub in our 2m for delta h Now this is where maths and physics would branch off: For maths: Leave your answer as sqrt(4g) For physics: Find the solution of sqrt(4g) with a calculator. We notice that the height is 2 m which is one significant figure and therefore we must leave our answer to 1 sf. ( g= 9.8 for gcse physics and 9.81 for a level or whichever the question wants you to use) we will use g=9.81 =>> v = 6.3 approx = 6

Hi! In this question, we can use the change in gravitational potential energy to calculate the kinetic energy that the car will have at the foot of the slope, and therefore it's velocity at this point. The mass of the car does not need to be known, as the mass in the 2 energy equations will cancel. 1.Change in GPE = m x g x change in h = 9.81x2xm =(19.62m) J 2. KE = change in GPE, since the car started at rest. So 1/2mv^2=19.62m The m's both represent the mass of the car, so can be divied through by on both sides. Leaving us with 1/2v^2 = 19.62 3. By rearranging this equation, we get v^2 = 2 x 19.62 v^2 = 39.24 v = 6.26418.. v = 6.26m/s