A child slides down a snow-covered slope on a sled. At the top of the hill, her

Question

A child slides down a snow-covered slope on a sled. At the top of the hill, her mother gives her a push at a speed of 1.3 m/s to get her started. The frictional force acting on the sled is one-fifth of the combined weight of the child and the sled. If she travels for a distance of 29.5 m and her speed at the bottom is 2.7 m/s, calculate the angle that the hill makes with the horizontal.

I want to make sure that I am doing this correctly:

(1.3)2+ (2*9.8*sin(theta)* 29.5) - ((2/5)*9.8*29.5)=2.72

Please let me know the correct answer and how to get there. Thanks!

Explanation / Answer

Wf = f*L = (1/5)*W*L = (1/5)*m*g*L

from energy conservation

PE + KEi - Wf = KEf

m*g*L*sintheta + 0.5*m*vo^2 - (1/5)*m*g*L = 0.5*m*vf^2

2*g*L*sin theta + vo^2 - (2/5)*g*l = vf^2

(2*9.8*29.5*sin theta) + (1.3*1.3)-((2/5)*9.8*29.5) = (2.7*2.7)

theta = 12.1 degrees

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