Respuesta :
Let the following denote:m - mass of the snowboarder, g - acceleration due to gravity, F - friction force, a - acceleration down the slope, R - normal reaction of the slope, u - coefficient of friction on the slope, v0 - snowboarder's initial speed, v - snowboarder's speed at the base of the slope, F1 - friction force on the level, a1- acceleration on the level,
Resolving parallel and perpendicular to the slope: mg sin(a) - F = ma would be equation 1 mg cos(a) = R would be equation 2F = uR would be equation 3
From (2) and (3): F = umg cos(a).
Substituting this in (1): mg[ sin(a) - u cos(a) ] = ma a = g[ sin(a) - u cos(a) ] = 9.81[ sin(28) - 0.18 cos(28) ] = 3.05 m/s^2.
v^2 = v0^2 + 2as v = sqrt(5.0^2 + 2 * 3.05 * 110) = 26.4 m/s.
F1 = - u1 mg - u1 mg = m a1 a1 = - u1 g = - 0.15 * 9.81 = - 1.47 m/s^2
Resolving parallel and perpendicular to the slope: mg sin(a) - F = ma would be equation 1 mg cos(a) = R would be equation 2F = uR would be equation 3
From (2) and (3): F = umg cos(a).
Substituting this in (1): mg[ sin(a) - u cos(a) ] = ma a = g[ sin(a) - u cos(a) ] = 9.81[ sin(28) - 0.18 cos(28) ] = 3.05 m/s^2.
v^2 = v0^2 + 2as v = sqrt(5.0^2 + 2 * 3.05 * 110) = 26.4 m/s.
F1 = - u1 mg - u1 mg = m a1 a1 = - u1 g = - 0.15 * 9.81 = - 1.47 m/s^2
Acceleration on the inclined plane and flat surface is different. The acceleration of snowboarding on the inclined plane is 3.05 m/sec² while on the flat surface is 1.4715 m/sec².
What is the friction force?
It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically it is defined as the product of the coefficient of friction and normal reaction.
(a)
On resolving the given force and accelertaion in the different components and balancing the equation gets.
Components in the x-direction
mgsina-F= ma
mgcosa=R
F=μR
F = μmgcosa
mg(sina-μcosa)=ma
a=g(sina-μcosa)
a=9.31(sin28°-0.18cos28°)
a= 3.05 m/sec²
Hence acceleration of snowboarding on the inclined plane is 3.05 m/sec²
(b)
According to Newton's third equation of motion;
v²=u²+2as
v²= (5)²+2×3.05×110
v=26.4 m/sec.
Fₓ×f= mgμ=ma
a=g×μ
a=9.81×0.15
a= 1.47 m/sec²
Hence acceleration on the flat surface is 1.4715 m/sec².
To know more about friction force refer to the link;
https://brainly.com/question/1714663
