Didn't I go through this with you before? Set initial KE equal to mu*mg*distance.
If you don't understand that, explain. I can't read minds.
A 61 kg skier on level snow coasts 184 m to stop from a speed of 12.0 m/s (A) use the work energy principle to find the coefficient of kinetic friction between the skis and the snow. B) suppose a 75 kg skier with twice the starting speed coasted the same distance before stopping, find the coefficent of kinetic friction between the skiers skis and the snow
6 answers
sorry about that...
I got your 1/2 m v^2=mu*mg*distance
so I did 1/2(61kg)(12.0 m.s)^2 = mu * (what do i put for mg) and distance is 184 m right?
I got your 1/2 m v^2=mu*mg*distance
so I did 1/2(61kg)(12.0 m.s)^2 = mu * (what do i put for mg) and distance is 184 m right?
m is mass, itis given as 61kg, g is 9.8m/s^2
so its the 61(9.8)
so then the final formula would look like
1/2(61kg)(12.0 m.s)^2 = mu * (61kg)(9.8m/s^2)(184m)
solve for mu right?
so then the final formula would look like
1/2(61kg)(12.0 m.s)^2 = mu * (61kg)(9.8m/s^2)(184m)
solve for mu right?
correct.
i did the left side of the equation = 4392
and the right side i got 109995.2
so i did 4392/109995.2 = .039929
and the right side i got 109995.2
so i did 4392/109995.2 = .039929