Clarification for Inclined Plane Q

[Hexon]

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from EK physics, q186, question states:

"2s is required for the mass to slide the distance d from a resting position. If the height h is increased by a factor of 4, while holding theta at 30 degrees, how long does it take the mass to slide the new distance d from rest?"

so i got as far as:
acceleration of block=gsin(theta)=(10)(sin30)=5m/s^2;
original distance for block=1/2*at^2=1/2*(5)(2)^2=10m

and since sine30=1/2=opposite/hypotenuse=h/d, i assumed the ratio between height (h) and distance (d) as being 1/2, so i reasoned that if u increased height by a factor of 4, then distance (d) would increased by a factor of 8.
therefore new distance=8d=8(10)=80m,

now solving for time:
displacement=1/2*at^2
80=1/2*(5)t^2
t=sqrt32=about 5ish. between 5 and 6, yet the answer choices were

a)4s
b)8s
c)16s
d)32s

the answer was a, but i'm just nary cuz the answer i got is midway between answer a) and b)

 

[UrshumMurshum]

This is how I solved it.

The ratio of sides for a 30 degree triangle is height : distance : length 1 : 2 : Sqrt(3)

So the ratio of height to distance is 1 : 2

The problem gives initial information that it takes 2s time units to travel 2 distance units (you can assume the initial height was 1 so the initial distance was 2)

Now you increase the height by a factor of 4 while keeping the angle constant.

In order to do this you must maintain the original triangle ratio

4 : 8 is the new ratio (you see how this maintains the 1 : 2 ratio of a 30 degree triangle?)

Your force stays the same.
F = mg * sin (theta)
F = mg h/d
Your height increased by a factor of 4, but so did your distance, so they cancel each other out and you're left with your initial force.

Now you need to solve for the initial acceleration this original force caused. Well the initial velocity is 0 and they give you distance and time so it's best to use this kinematic equation

2 = (1/2) * a * t^2
2 = (1/2) * a * 4(s)^2 Plugging in 2s for time
1/s^2 = a

Now solve for how long it takes with 8 distance units
8 = 1/2 * (1/s^2) * t^2
16 * s^2 = t^2
4s = t

 

[FishyTheFish]

The aceleration as you stated is 5m/s². You're also right about original distance of the block, 10m. This means the block needs to move down 10m along the ramp to get to the ground, however, we need to find the block's HEIGHT ABOVE the ground. 10sin30 = h and so h = 5.

Now we know that h is increased by 4. so now h = 4*5 = 20. This is the new HEIGHT ABOVE the ground of the block, and so the distance it needs to move along the ramp is d=20/sin30 = 40. Now using the kinematics.

deltaX = 1/2at²
40 = 1/2*5*t²
t=4

 

[Hexon]

thanks guys
such clean solutions lol; i guess it's easier once we have the number to plug n chug🙂

 

[MT Headed]

i assumed the ratio between height (h) and distance (d) as being 1/2, so i reasoned that if u increased height by a factor of 4, then distance (d) would increased by a factor of 8.

Umm, yeah. Please check your logic and dial again.

These kinds of mental shortcuts are the key to solving PS questions quickly, but you still gotta apply them correctly.

 

[minombre]

With questions like these you don't even need to take theta into account; just use the fact that y=1/2 a t^2, from kinematics. Then take the ratio between y2 and y1. You know that y2 is four times as big as y1, so:

y2/y1 = 4 = (1/2 * a2 * (t2)^2) / (1/2 *a1 * (t1)^2).

a is just g in both cases, and solving for t2:

t2 = t1 * sqrt(y2/y1) = (2sec) * sqrt(4) = 4 sec.

Taking ratios is a powerful tool because it eliminates dealing with the same constants over and over again and you only need to know relative values.

 

[FishyTheFish]

With questions like these you don't even need to take theta into account; just use the fact that y=1/2 a t^2, from kinematics. Then take the ratio between y2 and y1. You know that y2 is four times as big as y1, so:

y2/y1 = 4 = (1/2 * a2 * (t2)^2) / (1/2 *a1 * (t1)^2).

a is just g in both cases, and solving for t2:

t2 = t1 * sqrt(y2/y1) = (2sec) * sqrt(4) = 4 sec.

Taking ratios is a powerful tool because it eliminates dealing with the same constants over and over again and you only need to know relative values.

This is a very nice method! I'd just like to point out that a is not g, rather it is 5 m/s². But either way the a cancels so very nice method!

 

[Hexon]

With questions like these you don't even need to take theta into account; just use the fact that y=1/2 a t^2, from kinematics. Then take the ratio between y2 and y1. You know that y2 is four times as big as y1, so:

y2/y1 = 4 = (1/2 * a2 * (t2)^2) / (1/2 *a1 * (t1)^2).

a is just g in both cases, and solving for t2:

t2 = t1 * sqrt(y2/y1) = (2sec) * sqrt(4) = 4 sec.

Taking ratios is a powerful tool because it eliminates dealing with the same constants over and over again and you only need to know relative values.

Brilliant!
what a great shortcut!🙂
👍 to you, sir or madam😀