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Georgia [21]
3 years ago
6

Calculate the deceleration (in m/s2) of a snow boarder going up a 2.65° slope assuming the coefficient of friction for waxed woo

d 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
Physics
1 answer:
MAVERICK [17]3 years ago
8 0
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

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a)   v_average = 11 m / s, b)  t = 0.0627 s

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a) truck speed can be found with kinematics

         v² = v₀² - 2 a x

The fine speed zeroes them

           a = v₀² / 2x

           a = 22²/2 0.69

           a = 350.72 m / s²

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b) The average time

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             F = m a

             F = 2100 350.72

             F = 7.37 10⁵ N

.d) the ratio of this force to weight

             F / W = 7.37 10⁵ / (2100 9.8)

             F / W = 35.8

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- It is despised that the force is not constant in time

- Depreciation of materials deformation during the crash

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