Distance Question: 11kg Block Sliding Down 24° Frictionless Slope

[allora_97]

Homework Statement

A 11 kg block slides down a frictionless slope
which is at angle θ = 24◦. Starting from rest,
the time to slide down is t = 1.85 s.
The acceleration of gravity is 9.8 m/s2 .
(look at image)
What total distance s did the block slide?
Answer in units of m.

Homework Equations

F=ma
F=mg
tan(theta) = Fy/Fx

The Attempt at a Solution

I calculated the acceleration, a = F/m = 22.01116038m/s^2.
I used the time 1.85sec and the acceleration to find the velocity, v = 40.72064671m/s.
I don't know how to proceed from this point?

 

[aftershock]

Homework Statement

A 11 kg block slides down a frictionless slope
which is at angle θ = 24◦. Starting from rest,
the time to slide down is t = 1.85 s.
The acceleration of gravity is 9.8 m/s2 .
(look at image)
What total distance s did the block slide?
Answer in units of m.

Homework Equations

F=ma
F=mg
tan(theta) = Fy/Fx

The Attempt at a Solution

I calculated the acceleration, a = F/m = 22.01116038m/s^2.
I used the time 1.85sec and the acceleration to find the velocity, v = 40.72064671m/s.
I don't know how to proceed from this point?

What numbers did you use for F/m?

22 is way too high. It has to be under 9.8

 

[allora_97]

I found Fn = mg = 107.8
I used Force = Fn/tan24 = 242.1227642N
then i found a = F/m = 242.1227642N/ 11kg = 22.01116038 m/s^2

 

[Char. Limit]

I think you should have multiplied by tan(theta), not divided by it.

 

[allora_97]

thanks

 

[aftershock]

I found Fn = mg = 107.8
I used Force = Fn/tan24 = 242.1227642N
then i found a = F/m = 242.1227642N/ 11kg = 22.01116038 m/s^2

There's your mistake.

Multiply the force by sin(theta) then use that force in F/m to get your acceleration.