Help with block of ice on an incline, and friction

AI Thread Summary
An 8.80-kg block of ice slides down a 1.03m frictionless ramp, reaching a speed of 2.81m/s at the bottom. The angle of the ramp is calculated using the equation V^2 = V0^2 + 2ay, leading to an angle of approximately 22.85 degrees. The mass of the block does not affect the acceleration due to gravity, as both weight and inertia cancel each other out. For the second part of the problem, the impact of a constant friction force of 10.7 on the ice's speed at the bottom needs to be determined. The calculations for the speed under friction are still pending further assistance.
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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