Help with block of ice on an incline, and friction

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Homework Help Overview

The discussion revolves around a physics problem involving a block of ice sliding down an incline. The original poster seeks to determine the angle of the ramp and the speed of the ice when friction is considered. The context includes concepts of kinematics and dynamics, particularly focusing on forces, acceleration, and the effects of friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the angle of the ramp using kinematic equations but initially uses an incorrect ramp length. Some participants question the use of mass in the calculations, noting its cancellation in certain equations. Others provide insights into the relationship between mass, weight, and acceleration.

Discussion Status

The discussion has progressed with the original poster correcting their initial mistake regarding the ramp length. Some guidance has been offered regarding the calculations for acceleration and the relationship between forces. The original poster is now seeking assistance with the second part of the problem involving friction.

Contextual Notes

There is a specific mention of a constant friction force acting parallel to the ramp, which introduces additional complexity to the problem. The original poster is also uncertain about the correctness of their calculations and is looking for confirmation.

DrManhattanVB
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Homework Statement


A 8.80-kg block of ice, released from rest at the top of a 1.03m--long frictionless ramp, slides downhill, reaching a speed of 2.81m/s at the bottom.

need to find the angle between the ramp and horizontal, and what would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.7 parallel to the surface of the ramp?

Homework Equations



V^2=V0^2+2ay
F=mg or ma
frictional force =mu*F

The Attempt at a Solution


I tried to find the angle by using the equation :V^2=V0^2+2ay
with it being 2.8^2=(2)(9.8)(1.3)sin(x) with x being the angle, I got 17.9, which didnt work. also i don't see how the mass comes into play in this equation.

Thank you for any help
 
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Hi,

Are you sure that the problem is not just that you used 1.3 m for the length of the ramp when it is in fact 1.03 m long? The equation you have chosen does seem appropriate -- you want to find the acceleration given the distance traveled and the final velocity reached.

The mass doesn't come into play, since it has effectively already been canceled from both sides of the equation. Although more massive things have more inertia (in proportion to their mass) and are hence harder to acceleratate, they also have more weight (i.e. gravity pulls on them harder, also in proportion to their mass). These two effects cancel each other out, leading to the result that, for acceleration only under gravity, all objects will accelerate down the ramp at the same rate. This result will no longer be true once you throw in friction.
 
Thank you! i was using the wrong number!

can I get some help with the second part please?
 
2.82=2(x)(1.03)
7.84=2.06x
Divide by 2.06
x=3.806 m/s2 (acceleration)
mg(sin \theta) = ma
Divide by m..
g sin theta = a
9.8 sin \theta = 3.806
sin \theta = .3883
\theta = sin-1(.3883)
\theta = 22.852

Not sure if it's right, but I'm just going to guess?

EDIT: I'd also like to know if this is right lol
 
yes that is the correct answer. I am looking for help now on what would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.7 parallel to the surface of the ramp?
 

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