Physics Kinematics

[Maverick56]

A sled is moving down a frictionless slop inclined at an angle of 45 degrees w/ respect to the ground. If the length of the slope is 200 m, what is her velocity in m/s at the bottom, assuming that she starts from rest?

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[ThunderBear]

I'm not sure that this is correct since the final number is kind of weird, but I'm pretty sure the technique is correct.

For this problem we will use the convservation of mechanical energy. Starting from rest the object has no kinetic energy, and ends with a final kinetic energy. We will set the initial gravitational potential energy at the height of the slope, and arbitrarily specify that the final gratitational potential energy will be zero.

In other words initial gravitational energy = final kinetic energy. or...

mgh = 1/2mv^2

h is the height of the slope found through trig to be 2^(1/2)/2*200.

m cancels out.

solving for final velocity, v gives.

v = (g*2^(1/2)*200)^(1/2)

Which I can't simplify.

 

[amberkre]

I figured this one out using (Vf)^2= (Vo)^2+2(a)(d), but im not sure if my answer is correct:

Vo=0m/s
a=gsin(45)= 10(.707)= about 7 m/s^2
d=200m
Vf=?

(Vf)^2= (Vo)^2+2(a)(d)---> plug in the known and you get
(Vf)^2= (0)^2+2(7)(200)
(Vf)^2= 2(7)(200) = 2800 m/s
Vf= square root of (2800) = about 53 m/s

 

[BerkReviewTeach]

I'm not sure that this is correct since the final number is kind of weird, but I'm pretty sure the technique is correct.

For this problem we will use the convservation of mechanical energy. Starting from rest the object has no kinetic energy, and ends with a final kinetic energy. We will set the initial gravitational potential energy at the height of the slope, and arbitrarily specify that the final gratitational potential energy will be zero.

In other words initial gravitational energy = final kinetic energy. or...

mgh = 1/2mv^2

h is the height of the slope found through trig to be 2^(1/2)/2*200.

m cancels out.

solving for final velocity, v gives.

v = (g*2^(1/2)*200)^(1/2)

Which I can't simplify.

You have the right set up. The BR route would be the same thing, but with a shortcut applied. The height is about 200 m x 0.71, which is about 142 m.

Falling from a height of 142 m is between 125 m and 180 m, which would take between 5s and 6s, leading to a final speed between 50 m/s and 60 m/s. Because 142 is closer to 125 than 180, I'd put it in the ballpark of 53 m/s, which should be close enough to pick the best of four choices.