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A 75-kg snowboarder has an initial velocity of 5.0 m/s at the top of a 28 degree incline. After sliding down the 110-m long include (on which the coefficient of kinetic friction is u_k = 0.18), the snowboarder has attained a velocity V. The snowboarder then slides along a flat surface (on which u_k = 0.15) and comes to rest after a distance x.

Use Newton's second law to find the snowboarder's acceleration while on the incline and while on the flat surface.

A_incline = ?

A_flat = ?

Then use these accelerations to determine x.

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Added Tue, 16 Jun '15
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Use Newton's second law to find the snowboarder's acceleration while on the incline and while on the flat surface.

First - draw a Free Body Diagram.

Free Body Diagram of Snowboarder

Right will be positive X, and Up will be positive Y.

Sum of forces on the object when on the incline:

Fx_i = m*g_x - F_r = m*a

F_r = F_n * u_k.

Fy_i = F_n - m*g_y = 0

>> F_n = m*g_y

Fx_i = m*g_x - m*g_y*u_k = m*a

>> g_x - g_y*u_k = a

>> g*sin(28) - g*cos(28)*u_k = a.

a = 9.8 * sin(28) - 9.81 * cos(28) * 0.18

A_incline = 4.605 - 1.559 = 3.046 m/s^2

Sum of forces on the object when flat:

Fx_f = - F_r = m*a

F_r = F_n * u_k.

Fy_f = F_n - m*g = 0

>> F_n = m*g

Fx_f = -m * g * u_k = m*a

= -g * u_k = a

= 9.81 * 0.15

a = 1.4715 m/s^2

A_flat = 1.4715 m/s^2

Then use these accelerations to determine x.

For this part - we will need our kinematic equations.

We will first have to solve for the Velocity that the object reaches after the incline (V) and then solve for the distance that it goes afterwards.

recall :

V^2 = (V_0)^2 + 2*a*x

First solve for the speed at the bottom of the incline. We know that the incline is 110 meters long.

V^2 = 5^2 + 2 * 3.046 m/s^2 * 110 m

= 695.12 m^2/s^2

V = sqrt(695.12 m^2/s^2) = 26.36 m/s

Now solve for the distance with that as the initial speed.

0^2 = (25.88)^2 - 2 * 1.4715 m/s^2 * x

2 * 1.4715 m/s^2 * x = (25.88)^2

x = (26.36)^2 / (2 * 1.4715)

= 236.102 m

distance = 236.102 m

Edit
Added Tue, 16 Jun '15
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