A toboggan approaches a snowy hill moving at 11.0 m/s . The coefficients of stat

Question

A toboggan approaches a snowy hill moving at 11.0 m/s . The coefficients of static and kinetic friction between the snow and the toboggan are 0.420 and 0.320, respectively, and the hill slopes upward at 45.0 above the horizontal.

Part A

Find the acceleration of the toboggan as it is going up the hill.

Assume +x axis directed up the hill.

(answer is -9.15 m/s2, But i do not know how to get there)

Part B

Find the acceleration of the toboggan after it has reached its highest point and is sliding down the hill.

Assume +x axis directed up the hill.

Explanation / Answer

here,
velocity, v= 11 ms
coefficients fo friction are,
Static , us = 0.420
kinetic, uK = 0.320

Slope angle, A = 45 degrees

from newton second law, Fnet = ma

- mgSinA - uk*mg*CoA = ma -----------------(1)

solving for Acceleration, a
a = -gSinA -uk*gCosA
a = -9.8*Sin45 - 0.320*9.8*Cos45
a = -9.15 m/s^2

Part B:
When it slides down, eqn 1 can be written as
uk*CosA - mgSinA = ma

a = uk*gCoS - gSinA
a = 0.320* 9.8 * Cos45 - 9.8 * Sin45
a = -4.72 m/s^2

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