Calculate the deceleration (in m/s2) of a snow boarder going up a 2.65° slope assuming the coefficient of friction for waxed wood on wet snow. the equation a = g (sin (θ) - μk cos (θ)) for a snow boarder going downhill may be useful, but be careful to consider the fact that the snow boarder is going uphill. (enter the magnitude.) m/s2

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Physics

Allan Cuevas

28 April, 06:58

Calculate the deceleration (in m/s2) of a snow boarder going up a 2.65° slope assuming the coefficient of friction for waxed wood on wet snow. the equation a = g (sin (θ) - μk cos (θ)) for a snow boarder going downhill may be useful, but be careful to consider the fact that the snow boarder is going uphill. (enter the magnitude.) m/s2

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Ayaan Mcgee · Answer 1

The answer:

to solve this problem, we can use newton's second law

F=MxA

M is the mass of the object in motion

A is its acceleration (or deceleration)

F the sum of all exterior forces

in our case

the exterior forces are Fk (the friction force) and P (force of gravity)

In fact, Fk is positive because the motion is upward, its value is

Fk=μk MgcosΘ

and P=MgsinΘ

so we have F=MxA = μk MgcosΘ + MgsinΘ, and from where

A = μk gcosΘ + gsinΘ

μk=0.1 (coefficient of static friction on ice)

A=0.1x9.8xcos2.65° + 9.8xsin2.65° = 0.97 + 0.45 = 1.42 m/s²