Find the magnitude of acceleration of the puck on the ramp

AI Thread Summary
The problem involves calculating the acceleration of a 0.20 kg hockey puck launched up a 30-degree inclined metal ramp, with given coefficients of friction. The forces acting on the puck include gravitational force, normal force, and frictional force, which must be correctly resolved into components. The user initially calculated the normal force and friction but mixed up the sine and cosine components of gravity. The correct approach requires careful application of Newton's second law, ensuring that the forces are accurately represented in both the x and y directions. The user is encouraged to re-evaluate their calculations to find the correct acceleration value.
ScullyX51
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Homework Statement


A .20 kg hockey puck is launched up a metal ramp that is inclined at a 30.0 degree angle. The coefficients of static and kinetic friction between the hockey puck and the metal ramp are
Ms= .40 kg and Mk=.30 kg , respectively. The puck's initial speed is 35.0 m/s. What is the magnitude of the acceleration of the puck on the ramp?

Homework Equations


F= ma


The Attempt at a Solution


N= Nj
Fgrav= mg(cos 30- sin 30)
Friction= MkN

In the j direction the forces are:
N-mgsin(30)=0
N= -1.94
In the i direction the forces are:
mgcos(30) + MkN=ma
a= (mgcos(30)+ MkN)/ m =a

then when I plug in all of the given numbers I get -1.4.
I have tried this problem a bunch of times and cannot get anywhere close to the correct answer. What am I doing wrong?
 
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ScullyX51 said:

Homework Statement


A .20 kg hockey puck is launched up a metal ramp that is inclined at a 30.0 degree angle. The coefficients of static and kinetic friction between the hockey puck and the metal ramp are
Ms= .40 kg and Mk=.30 kg , respectively. The puck's initial speed is 35.0 m/s. What is the magnitude of the acceleration of the puck on the ramp?

Homework Equations


F= ma


The Attempt at a Solution


N= Nj
Fgrav= mg(cos 30- sin 30)
Friction= MkN

In the j direction the forces are:
N-mgsin(30)=0
N= -1.94
In the i direction the forces are:
mgcos(30) + MkN=ma
a= (mgcos(30)+ MkN)/ m =a

then when I plug in all of the given numbers I get -1.4.
I have tried this problem a bunch of times and cannot get anywhere close to the correct answer. What am I doing wrong?
mostly you've got your mgsin30 and mgcos 30 mixed up.
 
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