A child starts from rest and slides down a snow-covered hill with a slope angle of 47° from a height of 1.8m above the bottom of the hill. The speed of the child at the bottom of the hill is 3.1m/s. Find the coefficient of kinetic friction between the hill and the child.
My answer is 0.54, is that correct?
Potential energy at top = m g h = 1.8 m g
Kinetic energy at bottom = (1/2) m v^2 = .5 m(3.1)^2
Energy loss = L = m(1.8 g -.5*3.1^2)
work done by friction = L = F * d = F*1.8/sin 47
normal force = m g cos 47
F = mu * normal force = mu m g cos 47
so
work done by friction = mu m g cos 47 * 1.8 /sin 47
so in the end
mu g cos 47 * 1.8 /sin 47 = (1.8 g -.5*3.1^2)
I get closer to .78 but do not have calculator handy
Yes, that's what I'm getting too, thank you!
To find the coefficient of kinetic friction between the hill and the child, we can use the concept of conservation of mechanical energy.
First, let's calculate the potential energy (PE) of the child at the starting point. The formula for potential energy is given by:
PE = mass * gravity * height
where mass is the mass of the child, gravity is the acceleration due to gravity (approximately 9.8 m/s^2), and height is the vertical height from the top of the hill (1.8 m in this case).
Next, let's calculate the kinetic energy (KE) of the child at the bottom of the hill. The formula for kinetic energy is:
KE = (1/2) * mass * velocity^2
where mass is the same as before, and velocity is the speed of the child at the bottom of the hill (3.1 m/s in this case).
Since there is no significant loss or gain of energy due to other forces (such as air resistance), we can equate the potential energy at the top to the kinetic energy at the bottom:
PE = KE
mass * gravity * height = (1/2) * mass * velocity^2
Simplifying and canceling the mass on both sides of the equation, we get:
gravity * height = (1/2) * velocity^2
Now we can solve for the coefficient of kinetic friction (µ). The equation is:
µ = tan(θ) - (a/g)
where θ is the angle of the slope (47° in this case), a is the acceleration due to gravity, and g is the acceleration due to gravity.
Substituting the values:
µ = tan(47°) - (gravity / gravity)
Finally, we can substitute the value of the tangent of 47° into the equation:
µ = 1.072 - 1
µ ≈ 0.072
So, the coefficient of kinetic friction between the hill and the child is approximately 0.072.
Your answer of 0.54 is incorrect.