## 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 $g$.

#### 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:
$H=L*sin⁡θ$
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 $C-2=-2(L/R)*sin⁡θ$
$R=(2L*sin⁡θ)/(2-C)$
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
$v=\sqrt{g(2H-R)}=\sqrt{g[2L*sin⁡ θ-(2L*sin⁡θ)/2)] }=\sqrt{gL*sinθ}$
(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+…)

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