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?

Answers: 1

Show answers

Answers

answered: 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.

answered: Guest

answer would be letter choice (a), a ball resting on a shelf. would be another example of potential energy > stored energy > not moving.

a ball spinning on a finger > kinetic energy > moving energy

a ball rolling on the ground > kinetic energy > moving energy

a ball flying through the air > kinetic energy > moving energy

hope that : )

answered: Guest

Did you ever figue this

Other questions on the subject: Physics

Physics, 22.06.2019 09:00, cmflores3245

Sobre transformações cíclicas, são feitas algumas afirmações, determine quais estão corretas. i – a variação de energia interna em uma transformação cíclica é nula; ii – o trabalho, em uma transformação cíclica, é sempre positivo; iii – ao completar um ciclo, o gás tem a mesma temperatura com a qual começou a transformação; iv – uma transformação cíclica é, obrigatoriamente, composta de transformações conhecidas, como adiabática, isocórica, etc: a) i e iii b) i e ii c) ii e iii d) i e iv e) iii e iv

Answers: 1

continue

Physics, 22.06.2019 15:30, bloodry

Two point charges are separated by a certain distance. how does the strength of the electric field produced by the first charge, at the position of the second charge, change if the second charge is doubled? a. the field does not change b. the field strength decreases by half. c. the field strength doubles d. the field strength quadruples

Answers: 1

continue

Physics, 22.06.2019 16:00, cmir

The electric field direction is defined by the direction of the force felt by (select one of the following answers): 1. a negative charge. 2. a positive charge. 3. both positive and negative charges.

Answers: 2

continue

Physics, 22.06.2019 18:00, emory238

What structural formula represents 4 electrons shared between two atoms?

Answers: 1

continue

Do you know the correct answer?

A200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. the coefficients of...