Will you answer change if the stone of the hill changes, but it remain the same

Question

Will you answer change if the stone of the hill changes, but it remain the same height? if the height of the hill changes, but it remains the same heights? If the height of the hill changes? Explain. (c) If the frictional force is 2.5 N, what is the speed of the sled when it reaches the bottom of the hill? (d) if the sled now travels on horizontal ground at the speed found in (c) and encounters the same frictional force, how far can the sled travel before coming to stop? (e) What happens to the sled's Kinetic energy in (d)? Explain how this differs from a sled coming to rest as it travels up a hill.

Explanation / Answer

Given mass of sled m = 2.5 kg, height h = 5 m, the slope is theta = 10 degrees

when th sled starts from rest , the forces acting on the sled is the force of gravity (verticle) associated potential energy

   F = mg sin theta

   W = mg sin theta *h

the gravitaional potentila energy is p.e = mgh sin theta

will be converted in to kinetic energy completely at the bottom so the speed of the sled when it reaches the bottom of the hill is

a)
   mgh sin theta = 0.5*mv^2

   v^2 = (2gh sin theta)

   v = sqrt(2gh sin theta)

   v = sqrt(2*9.8*5 sin 10) m/s
   v = 4.1252 m/s

b) as the speed is proportional to both the slope and the height

so if the slope changes , remains height constant andn

   speed can also changes with changing the height.

c) with frictional force Ff = 2.5 N

then

   mgh sin theta- Ff = 0.5*mv^2

   v = 2(gh sin theta-Ff)

   v = sqrt(2(gh sin theta- Ff))

   v = sqrt(2(9.8*5 sin 10 - 2.5)) m/s
  
   v = 3.4667 m/s

d) work done by the frictional force causes the sled to come to rest

   Ff*s = change in k.e = 0.5*m(v2^-v1^2)

   s = 0.5*2.5(0^2-3.4667^2)/2.5 m

   s = -6.00900 m

e) when the sled travelling up the hill the forces acting on it are to the opposite direction to the motionn of the sled is

   the frictional force and the gravitaionnal force component , so it comes to rest within a short distance
but on the surface only the frictional force will act on the sled so that it come to rest after a longer displacement
  

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