Ramp/Mass question?

Question

For some reason, I just cannot get this problem. Well, here is the scenario:

A 20kg block is placed on a ramp of 22degrees. The block has been shot at a speed of 3 m/s from the bottom of the ramp. Assuming that the ramp is frictionless, how far up will the block go? And how long will it take for the block to return back to the "shooting" point?

Here is a picture of what the scenario looks like:

http://i91.photobucket.com/albums/k285/happygolucky2442/Aphunker.jpg

Thanks in advance, best answer to the person with the best explanation!

Answer 1

some of the weigh of the block is taken down the ram and some of its weigh is taken perpendicular to the ramp.

find the x-component of its weigh.
Fweigh=mg
Fweigh=(20kg)(9.8)
Fweigh= 196N

Fw(x) = sin(22)(196N=73.42N


a=Fnet/m
since there is no friction, Fw(x) is Fnet
a= Fw(x) / m
a= -73.42N / 20kg
a= -3.671m/s^2

Vf^2=2ad+Vi^2
at the maximiun distance, its final velocity is 0m/s
(0m/s)^2 = 2(-3.671m/s^2)d + (3m/s)^2
-9m^2/s^2 = (-7.342m/s^2)d
d= 1.225m

when it return to its shooting point, its final speed is the same as the speed it leaves, except it's negative because we are dealing with velocity

Vf=at+Vi
-3m/s = (-3.671m/s^2)t+3m/s
-6m/s = (-3.671m/s^2)t
t= 1.63s

check my calculation.

Answer 2

Im no rocket scientist,and I'm dumber than a football player, but from what I know, you are working against gravity. So what if its shot up a ramp? It still has to work against gravity. How long would it take for a 20kg block to go straight up and striaght down?
Think of it like this, No friction. When a Punter kicks a football, the hang time is relevant to the height. Suppose a Punter kicks a 20kg football at 22degrees at a rate of 3m/s. How far will it travel and whats the hang time? Hike..

Answer 3

It's pretty easy actually: treat it exactly like a vertically launched free-falling object, but with the gravitational acceleration reduced by a factor which equals the sine of the angle of the ramp from the horizontal.

Av = 9.81(m/s^2) * sin(22°)

You can completely ignore the horizontal component of the motion. It's irrelevant to the total time.

Answer 4

Since there is no friction, the problem is simple.

Find the vertical height and time as if the object is thrown vertically (with out using the ramp)

Using the formula V^2 -U^2 = 2as, v=0 and a = -g

H = (U^2 / 2g) = 3x3 / (2x 9.8) = 0.459 m.

Using the figure we measure the length along the plane for this vertical height.

Sin 22 = H/L.

L = 0.459 / sin 22 = 1.226 m.
------------------------------------------------------------------------
To find the time for the object to return to its starting place,

u = 3 and v = -3.

t = (v - u) / a = 6/9.8 = 0.612 s.
----------------------------------------------------------------------

Another method.

The time taken for the object to go up through the height 0.459 m is found using

v = u + a t

0 = 3 - 9.8 t1

t1 = 0.306 s
To return it takes the same time.

Total time is 2 x 0.306 = 0.612 s.

But the same time is found using a simple method given above.
------------------------------------------------------------------------------------
Another way of finding the distance.


From the figure the force pulling down and parallel to the plane down the 20kg mass is 73.4 N.

Down ward acceleration parallel to the plane is F/m

= 73.4/ 20 = 3.67m/s^2

Initial speed = 3m/s.
Final speed = 0m/s.
Acceleration = - 3.67m/s^2 (opposite to 3m/s.)
Displacement = L.

Using v^2 - u^2 = 2as

0 - 3x3 = - 2 x 3.67 x L
L = 9/ (2 x3.67) = 1.23m.
--------------------------------------------------------------------

Answer 5

vertical component of 3 m/s = 3*sin 22 = 1.123 m/s
horiz componenet of 3 m/s = 3*cos 22 = 2.782 m/s

Calculate time it takes to reach top of trajectory:
v(final) = v(orig.) + a*t
0 = 1.123 + -9.8*t
t = -1.123/-9.8
t = 0.115 seconds

Calculate the distance it travels to top of trajectory:
d = v(orig.)*t + 0.5*a*t^2
d = 1.123*0.115 + 0.5*-9.8*0.115^2
d = 0.129 - 4.9*0.013
d = 0.129 - 0.0637
d = 0.0653 metres

Now that we've calculated the distance it has travelled vertically we have to translate this back to parallel to the ramp.

sin 22 = 0.0653/x
x = 0.0653/sin 22
x = 0.1743 metres up the ramp

The total time is double the time it took to get to the top of it's trajectory:
2*0.115 = 0.23 seconds to return to start.

Answer 6

first find out how fast it will move up. do (3)sin22, which equals 1.12m/s. then use formula vf^2=vi^2+2as to get distance up. turns out to be .064m using -9.8 as a. then do .064/sin22 to get distance up the ramp. turns out to be .17m. then find acceleration up the ramp. use same formula again. 0=9+2a(.17). a turns out to be -26.47. then use vf=vi+at to get t. turns out to be .34seconds