🚀 Physics

# 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?

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

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