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camjam73
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Golf Cart Questions
Ms. Boylan on her 1750-kg cart is traveling at a constant speed of 8.5-m/s on a wet, straight, and slippery road. A mindless student steps out in from of the cart. The wheels lock and she begins skidding and slides to a halt in 12.0-m. What is the coefficient of sliding friction between the tires and the slippery road?
vf^2 = vi^2 + 2ad
a = (Fapplied - Fopposed) / mass
u = Ffriction / weight
Her acceleration is found with:
(0)^2 = (8.5)^2 + 2(a)(12)
a = -3.01 m/s^2
So, -3.01 equals applied force minus Ffriction over 1750.
Then I get stuck. If I could just figure that out, I could divide it by (1750 * 9.80) to get the coefficient. Thanks for your help.
Homework Statement
Ms. Boylan on her 1750-kg cart is traveling at a constant speed of 8.5-m/s on a wet, straight, and slippery road. A mindless student steps out in from of the cart. The wheels lock and she begins skidding and slides to a halt in 12.0-m. What is the coefficient of sliding friction between the tires and the slippery road?
Homework Equations
vf^2 = vi^2 + 2ad
a = (Fapplied - Fopposed) / mass
u = Ffriction / weight
The Attempt at a Solution
Her acceleration is found with:
(0)^2 = (8.5)^2 + 2(a)(12)
a = -3.01 m/s^2
So, -3.01 equals applied force minus Ffriction over 1750.
Then I get stuck. If I could just figure that out, I could divide it by (1750 * 9.80) to get the coefficient. Thanks for your help.
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