Question

A child on a sled is perched at the top of a hill, at a height 7.0 m above the base of the hill. The total mass of the girl plus the sled is 32 kg. Ignore friction and air resistance. If the child/sled starts sliding down the hill with an initial speed of zero, how fast will she be moving when she reaches the base of the hill? If her mother gives her a strong push so that she starts down the hill with an initial speed of 5 m/s instead of zero, how fast will she be moving when she reaches the base of the hill? show all work

Answer 1

a)11,71 m/s

b)12.73 m/s

Step-by-step explanation:

We apply the conservation principle because there is no friction between the hill and the sled:

Total initial energy (Ei) = Total initial energy (Ef), Equation(1)

We define:

K = Kinetic energy =
`(1)/(2)*m*v^(2)`

U = Potential energy =
`m*g*h`

m = mass (kg)

v = velocity (m/s)

h = height (m)

g = gravity acceleration =
`9.8m/s^(2)`

A) hi=7m, hf=0 ,vi=0 , vf=?

`Ei=Ef\\Ki+Ui=Kf+Uf`

`1/2*m*vi^(2) +m*g*hi= 1/2*m*vf^(2) +m*g*hf`

`0+m*g*hi=1/2*m*vf^(2)+0`

`(2*m*g*hi)/m=vf^(2)`

`vf=√(2*g*hi)`

`vf=√(2*9.8*7)`

`vf=11.71 m/s`

Answer: She reaches the base of the hill with speed of 11.22 m/s

B) hi=7 m, vi=5 m/s, hf=0, vf=?

`m*g*hi+1/2*m*vi^(2) =m*g*hf+1/2*m*vf^(2)`

We divide all terms by m

`g*hi+1/2vi^(2) =g*hf+1/2*vf^(2)`

`9.8*7+1/2*5^(2) =1/2*vf^(2)`

`2(68.6+12.5)=vf^(2)`

`162.2=vf^(2)`

`vf=√(162.2)`

`vf=12.73 m/s`

Answer: She will move with a speed of 12.73 m/s when she reaches the base of the hill