Finding a new speed after adding a friction force? Help

[monikraw]

Homework Statement

A 8.00kg block of ice, released from rest at the top of a 1.04m--long frictionless ramp, slides downhill, reaching a speed of 2.76m/s at the bottom.

What is the angle between the ramp and the horizontal?
This I calculated as 21.9 degrees

What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.9N parallel to the surface of the ramp?

Homework Equations

F = ma
w = mg
vx = v0 + axt

i might be missing several

The Attempt at a Solution

so I drew a diagram which i probably can't move to here, but--
I aligned the x and y-axis to match the normal force and the friction force

I calculated the force of the block of ice going down the ramp by using "mgsin(theta)" (where theta is 21.9) and got 29.24 N. I subtracted 10.9N from it to get the net force of 18.34N. I used F = ma to find acceleration and I got 2.293m/s^2.

I tried putting in 2.293m/s^2 and it says close but not quite.
How do I find t so I can use the vx = v0 + axt?

I got stuck up to there and some(or all of it) may be wrong since physics is my weakest subject haha.

 

[CAF123]

Homework Statement

A 8.00kg block of ice, released from rest at the top of a 1.04m--long frictionless ramp, slides downhill, reaching a speed of 2.76m/s at the bottom.

What is the angle between the ramp and the horizontal?
This I calculated as 21.9 degrees

Correct

What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.9N parallel to the surface of the ramp?

Homework Equations

F = ma
w = mg
vx = v0 + axt

i might be missing several

The Attempt at a Solution

so I drew a diagram which i probably can't move to here, but--
I aligned the x and y-axis to match the normal force and the friction force

I calculated the force of the block of ice going down the ramp by using "mgsin(theta)" (where theta is 21.9) and got 29.24 N. I subtracted 10.9N from it to get the net force of 18.34N. I used F = ma to find acceleration and I got 2.293m/s^2.

I tried putting in 2.293m/s^2 and it says close but not quite.
How do I find t so I can use the vx = v0 + axt?

I got stuck up to there and some(or all of it) may be wrong since physics is my weakest subject haha.

Why not use $v^2 = v_{o}^2 + 2as ?$ Could also use energy methods.

 

[monikraw]

I got the answer after dividing by 1.04. I don't know why that happened to be the answer nor do I know why that was a necessary step. Someone please explain?

 

[CAF123]

I got the answer after dividing by 1.04. I don't know why that happened to be the answer nor do I know why that was a necessary step. Someone please explain?

That would not make any sense. Dividing an acceleration by a length does not give a speed. (Check the dimensions)
Use what I suggested in my previous post.

 

[monikraw]

where s is seconds, I'm assuming?
so v^2 = 0 + 2*(2.293m/s^2)(0.377m/s?)
I got time by dividing 2.76 by 1.04 since d = speed * time
square root each side and I get v = 1.314556188

the answer was 2.19 :x...probably did something wrong.

 

[CAF123]

where s is seconds, I'm assuming?

The $s$ in this eqn is the displacement of the body. The value of $s$ here is the 1.04m, since we are looking for the speed at the bottom of the slope and the body moved 1.04m from where it was released.

so v^2 = 0 + 2*(2.293m/s^2)(0.377m/s?)

Check this again, given what I said above.

I got time by dividing 2.76 by 1.04 since d = speed * time

From the moment the body is released, it will gain speed. The eqn d=vt only applies if the body is not accelerating.