Question

Two bicycle tires are set rolling with the same initial speed of 3.00m/s along a long, straight road, and the distance each travels before its speed is reduced by half is measured. One tire is inflated to a pressure of 40 psi and goes a distance of 18.1m ; the other is at 105 psi and goes a distance of 93.2m . Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be g = 9.80m/s2 .

What is the coefficient of rolling friction μr for the tire under low pressure?

I tried solving F=ma= mu Force normal

ma=umg. mass cancels. a=ug

v(f)^2 = v(i)^2 + a(x(f)-x(i))
1.5^2=3^2+u(18.1)(9.8).

I solved for u and got 0.03805 and it was wrong.

Answer 1

You used the right equation, but you forgot the 2. Its
v(f)^2 = v(i)^2+2u(18.1)(9.8)

Answer 2

-0.0190269475

Answer 3

0.0243544932

Answer 4

assdasdasd

Answer 5

I don't like this question