Posted by **Mary** on Saturday, February 11, 2012 at 10:47pm.

A child insists on going sledding on a barely snow-covered hill. The child starts

from rest at the top of the 60 m long hill which is inclined at an angle of 30o to the horizontal, and arrives at the bottom 8.0 s later. What is the coefficient of kinetic friction between the hill and the sled?

- Physics -
**drwls**, Saturday, February 11, 2012 at 11:47pm
Assume that the acceleration is constant, such that

60 m = (a/2)t^2

a = 120/(8)^2 = 1.875 m/s^2

Then use Newton's second law in the form

Fnet = m*g sin30 - m*g*cos30*u = m*a

Note that the mass cancels out, which is good since they did not provide a value.

With the value of a that you now know, the friction coefficient u can be calculated.

0.50 g - 0.866 u*g = 1.875 m/s^2

8.49 u = 4.90 -1.875 = 3.02 m/s^2

u = 0.356

## Answer This Question

## Related Questions

- Physics-1 - A child sleds down a snow covered hill with a uniform acceleration. ...
- Physics - A small child is learning to ride a bike for the first time. Her dad ...
- AP Physics - A child on a sled starts from rest at the top of a hill and slides ...
- physics - A child holds a sled on a frictionless, snow- covered hill, inclined ...
- physics - A child holds a sled on a frictionless, snow- covered hill, inclined ...
- Physics - A child holds a sled on a frictionless, snow- covered hill, inclined ...
- physics - The cart starts from rest at the top of a hill with height, H2. It ...
- physics - The cart starts from rest at the top of a hill with height, H1. It ...
- Physics - A skier starts from rest at the top of a hill that is inclined at 9.8...
- physics - A skier starts from rest at the top of a hill that is inclined at 10.3...

More Related Questions