Question

At the base of a frictionless icy hill that rises at 25.0∘ above the horizontal, a toboggan has a speed of 11.9 m/s toward the hill. How high vertically above the base will it go before stopping?

Answer 1

Answer:

17.10 m

Explanation:

When the toboggan stops, we have:

$N-w_y=0\\F-w_x=0$

The x-component of weight is the product of the weight and the sine of the angle above the horizontal, so the y-component of W is the product of the weight and the cosine of the angle above the horizontal.

$N-mgcos(25^\circ)=0\\F-mgsin(25^\circ)=0\\F=mgsin(25^\circ)(1)$

The work-energy principle states that the change in the kinetic energy of an object is equal to the net work done on the object.

$\Delta K=W\\K_f-K_i=F\cdot h\\\frac{m(v_f)^2}{2}-\frac{m(v_i)^2}{2}=Fhcos(180^\circ)$

Recall that $v_f=0$, replacing (1):

$-\frac{m(v_i)^2}{2}=mgsin(25^\circ)h(-1)\\h=\frac{(v_i)^2}{2gsin(25^\circ)}\\h=\frac{(11.9\frac{m}{s})^2}{2(9,8\frac{m}{s^2})sin(25^\circ)}\\h=17.10 m$