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

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An 11 kg block slides down a frictionless slope at a 24° angle, taking 1.85 seconds to reach the bottom. The initial calculations for acceleration were incorrect, as the force should be adjusted using the sine of the angle rather than the tangent. The correct approach involves calculating the gravitational force component acting down the slope, which is mg * sin(θ). This will yield a proper acceleration value under 9.8 m/s². The discussion emphasizes the importance of using the correct trigonometric function to determine the block's acceleration and subsequent distance traveled.
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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?
 

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allora_97 said:

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
 
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
 
I think you should have multiplied by tan(theta), not divided by it.
 
thanks
 
allora_97 said:
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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