A skier starts from rest at the top of a hill that is inclined at 12.0° with the horizontal. The hillside is 1?

First find the forces acting on the skier as he moves down the hill. There are two: gravity and friction.

Because the skier is on an incline, only a component of gravity is accelerating the skier down the hill:

(mass x gravity x sin12)

The force of friction (FF) opposes the motion of an object. On a horizontal surface, the force of friction is equal to Uk x n, where (n) is the normal force and is equal to (mass x gravity). On this incline the normal force is equal to:

(mass x gravity x cos12)

So the frictional force is equal to:

Uk x m x g x cos12

We can now solve for the total net force acting on the skier as he moves down the hill. The net force (F) is equal to:

F = m x a = (m x g x sin12) - (Uk x m x g x cos12)

Mass cancels

a = (g x sin12) - (Uk x g x cos12)

a = 1.32 m/s^2

Now solve for the Velocity (V) at the bottom of the hill using this equation.

V^2 = Vo^2 + (2 x a x d)

Vo = 0 (start from rest)

a = accel (1.32m/s^2)

d = distance (190m)

V = square root (2 x a x d)

V = square root (2 x 1.32 x 190) = 22.4m/s

This is the velocity of the skier as he levels out at the bottom of the hill. At this point the only force causing a net acceleration is friction. On a horizontal surface, the frictional force is equal to:

Uk x m x g

So:

Uk x m x g = m x a

Solve for the new acceleration (a)

a = Uk x g

a = .0750 x 9.80 = -.735m/s^2

Now solve for distance:

V^2 = Vo^2 - (2 x a x d)

V= zero

Vo = 22.4m/s

a =.735m/s^2

Solve for d

d = (Vo^2 / 2 x a)

d = (22.4)^2 / (2 x .735) = 341m

Hope this helps.

WELL, its been about 3 months since I've done anything Physics related... but here goes:

many parts but still simple: how long will he accelerate down the hill at what rate of acceleration and then his velocity at the bottom of the slope then we can find how long he keeps moving.

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the skier will be accelerating at a=(sin(12)-.0750)*9.8 meters per second squared down the hill

the hill is 190 meters long and D=.5*a*t^2+Vi*t but we can use D=.5*a*t^2 since he starts at rest

solve for t we get t=sqrt(2*d/a) and we plug in known values and get t=sqrt((2*190)/9.8*(sin(12)-.075)

so he will accelerate for t seconds at a m/s^2 we can use V=a*t... (this will turn into an ungodly mess without math templates so ill just use the variables) and now we know he's going V m/s at the bottom and then he will decelerate at a -.075*9.8 m/s^2 we use Vf^2=0=Vi^2+2*a*d, solve for d and get Vi^2/-2a=d and then we can say from that V^2/2*.075*9.8 I'll use my trusty Ti-Nspire to calculate the distance and I get... 336.71 meters for the final distance, but be sure to double check, I always make stupid little mistakes like not squaring something or something like that.... But, if I was taking a test I would be comfortable with that number

properly, its been approximately 3 months considering i've got finished something Physics appropriate... yet right here is going: many aspects yet nevertheless elementary: how long will he develop up down the hill at what fee of acceleration and then his velocity on the backside of the slope then we are able to discover how long he retains shifting. --- the skier would be accelerating at a=(sin(12)-.0750)*9.8 meters in keeping with 2nd squared down the hill the hill is a hundred ninety meters long and D=.5*a*t^2+Vi*t yet we are able to apply D=.5*a*t^2 considering he starts at relax resolve for t we get t=sqrt(2*d/a) and we plug in known values and get t=sqrt((2*a hundred ninety)/9.8*(sin(12)-.0.5) so he will develop up for t seconds at a m/s^2 we are able to apply V=a*t... (this could become an ungodly mess without math templates so sick purely use the variables) and now all of us be attentive to he's going V m/s on the backside and then he will decelerate at a -.0.5*9.8 m/s^2 we use Vf^2=0=Vi^2+2*a*d, resolve for d and get Vi^2/-2a=d and then we are able to declare from that V^2/2*.0.5*9.8 i will use my trusty Ti-Nspire to calculate the gap and that i'm getting... 336.seventy one meters for the purely suitable distance, yet verify to double examine, I constantly make stupid little errors like no longer squaring something or something like that.... yet, if i replaced into taking a attempt i could be gentle with that style

dk