Vector ramp question - time taken for object to slide down ramp

AI Thread Summary
A 9.0 kg box is sliding down a ramp at a 24-degree angle with a coefficient of kinetic friction of 0.25. The net force acting on the box was calculated to be 15.7 N after determining the gravitational and frictional forces. To find the time taken to slide 2.0 m, the user initially struggled with the correct application of kinematic equations. Eventually, using the equation d = Vot + 0.5at² led to the correct time of 1.5 seconds. The discussion highlights the importance of understanding net force and kinematics in solving motion problems.
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Homework Statement



A 9.0 kg box is sliding down a smooth flat ramp which makes an angle of 24 degrees with the horizontal. If the coefficient of kinetic friction (μ) between the box and the ramp is 0.25, how long will it take the box to slide 2.0 m down the ramp froma standing start?


Homework Equations



Fg = mg
Ff = μFn
SOH/CAH/TOA


The Attempt at a Solution



Fg= 9.0 x 9.8 =88.2 N
From a vector diagram, I solved Fn using cos(24)x88.2= 80.6 N
Therefore, Ff = 0.24 (80.6) = 20.2 N
Fd (force down the ramp parallel to ramp) = sin(24) x 88.2 = 35.9

Fnet = 35.9 - 20.2 = 15.7 N

Now to solve d=vt.

d= 2.0m

but do i use 15.7 as my velocity or is it acceleration and if so, how should I approach the calculus? Or is the velocity also 15.7 m/s because it starts at rest. If so, i divide 2 by 15.7 and get a 0.13 but the answer in my textbook is 1.5

Thanks for your help :smile:
 
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avsj said:

Homework Statement



A 9.0 kg box is sliding down a smooth flat ramp which makes an angle of 24 degrees with the horizontal. If the coefficient of kinetic friction (μ) between the box and the ramp is 0.25, how long will it take the box to slide 2.0 m down the ramp froma standing start?


Homework Equations



Fg = mg
Ff = μFn
SOH/CAH/TOA


The Attempt at a Solution



Fg= 9.0 x 9.8 =88.2 N
From a vector diagram, I solved Fn using cos(24)x88.2= 80.6 N
Therefore, Ff = 0.24 (80.6) = 20.2 N
Fd (force down the ramp parallel to ramp) = sin(24) x 88.2 = 35.9

Fnet = 35.9 - 20.2 = 15.7 N

Now to solve d=vt.

d= 2.0m

Everything up until here is correct. What you have is the net force being experienced by the block. Now, you know F=ma, so to find the acceleration you need to divide the force by the weight of the object.

Then you can solve for time using the acceleration and distance with equations of kinematics.
 
Thanks a lot :D, ... but I haven't learned the equations of kinematics yet. Could you please tell me which ones are relevant?

Kinematics is our next unit, this is a 'brain buster' at the end of my current unit.

Thanks so much
 
F= ma

15.7/9 = 1.7 acceleration

I tried d= (Vo + Vf)/2 x t

so 2 = 1.7/2 x t

t = 2.4

The correct answer is 1.5
 
Oooh,

Should I be using d= Vot + .5at^2 ?

This gives me the correct answer. Great thanks chaoseverlasting :D
 
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