# Icy slope and sled

#### The icy slope

A frictionless icy slope of angle $\theta$ having a length $L$, followed by a horizontal portion, then by a bump ($R$) make a frictionless path as in the image. The weight of the sled on this path is C*TrueWeight when the sled is exactly at the bump’s top, ($0 < = C < 1$).

a) Please find the value of R as function of $C$, $\theta$ and $L$.

b) Find $R=?$ when $\theta=26 deg$, $C = 0.75$, and $L = 5 m$.

c) When $C = 0$, the sled jumps at the top of the bump. What is the horizontal distance that the sled travels as a function of $L$, $R$, $\theta$ and $g$.

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#### The bump

At the top of the bump the total force is the difference between the sled weight and the centrifugal force. The centrifugal force acts outside the path.

$C*(mg)=mg-F_cf=mg-(mv^2)/R$

From the energy conservation considerations we write the equation:

$mgH=(mv^2)/2+mgR so mv^2=mg(2H-R)$

where we find the initial height from the equation:

$H=L*sinθ$

Thus the dependence of $R$ on the $L$, $\theta$ and $R$ is

$C(mg)=mg-mg/R*[2L*sin (θ-R)]$

$C=1-1/R*(2L*sin (θ-R))$ or $C=1-2*(L/R)*sinθ+1$ or $C-2=-2(L/R)*sinθ$

$R=(2L*sinθ)/(2-C)$

b)

With the values from text we find for $R$,

$R=(2*5*sin 26)/(2-0.75)=3.50 m$

#### Range of sled

c)

At the top of the bump the speed is

$v=\sqrt{g(2H-R)}=\sqrt{g[2L*sin θ-(2L*sinθ)/2)] }=\sqrt{gL*sinθ}$

(We can apply also energy considerations to find this.)

Thus, the time to fall from height R (on top of bump) is

$R=(gt^2)/2$ so $t=\sqrt(2R/g)$

Finally, the range of the sled after it jumps over the top of the bump is the product of the speed with the time:

$X=v*t=\sqrt{gL*\sin θ*(2R/g)}=\sqrt{2LR*\sin θ }$

Observation: On all frictionless paths, one can apply the principle of conservation of energy to find the unknown values (for example the final speed). (The initial potential energy = Final kinetic energy+…)