Physics, 05.08.2019 01:00 vittoriochavez9700

A200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. the coefficients of static and kinetic friction between the hockey puck and the metal ramp are μs = 0.40 and μk = 0.30, respectively. the puck's initial speed is 90 m/s. what vertical height does the puck reach above its starting point?

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Answer from: maxicanofb0011

The vertical height above its starting point will be h=860.28m

Explanation:

The first thing before doing our calculations, will be to put the mass of the hockey puck in kg instead of g, to be consistent with the units.

The second thing to notice is that we have μs and μk, but we will use μk to calculate our answer because the hockey puck will be in motion until it stops. The fact that μs is greater than μk, tells us that when the it stops, it will remain in that place and not move again.

To resolve the problem, we have that μk=0.3, v₀=90m/s, m=0.2kg, θ=30°, and put a system of coordinates inclined 30° (to coincide x-axis with the surface of movement).

Now, we calculate Newton's equations:

$N-mgcos\theta=0$

for the y-axis, and

$-mgsin\theta-F\mu=ma$

for the x-axis. In this equations N is the normal force, g is the gravity, a is the puck acceleration and Fμ is the frictional force, wich can be written as

$F\mu=\mu_{k}N=\mu_{k}mgcos\theta$

using the first equation... then from the second equation, we clear a, getting

$a=-gsin\theta-\mu_{k}gcos\theta$

Until here, we used Newton's equations, and now we go to the kinematic equations, knowing that an acceleration is a derivative of a velocity, and a velocity is a derivative of a position.

With this in mind, we integrate

$\int\limits^v_{v_{0}} \, dv=\int\limits^t_0 ({-gsin\theta-g\mu_{k}cos \theta}) \, dt$

from where we get equation A

$v=v_{0}-gt(sin\theta-\mu_{k}cos\theta)$

Now, we integrate for the second time, and get that

$\int\limits^x_{0} \, dx=\int\limits^t_0 ({v_{0}-gt(sin\theta-\mu_{k}cos \theta})) \, dt$

to obtain equation B

$x=v_{0}t-\frac{g}{2}t^2(sin\theta-\mu_{k}cos\theta)$

Then, we equalize equation A to zero to get the time when the puck stops (v=0), and clear t to put it into equation B, this will give us the distance that the puck travels:

$v=0 \Leftrightarrow t=\frac{v_{0}}{g(sin\theta-\mu_{k}cos\theta)}$

and

$x(v=0)=\frac{1}{2}\frac{v_{0}^2}{g(sin\theta-\mu_{k}cos\theta)}$

therefore, we can replace the given data and calculate

x=1720.56m

Finally, in order to know the vertical height, we use trigonometrics, and calculate the following relation

$h=(x)sin\theta=(1720.56m)sin(30)=860.28m$

wich is the answer to the problem.

Answer from: Quest

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Answer from: Quest

there are no contact forces mentioned on the list of choices that you've provided.

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