Tough physics question, need help!?

Question

A 4 Kg block is placed at rest at a vertical height of 21 meters on a plane inclined at 5o with the horizontal. If the surface is frictionless, how fast will the block be moving when it gets to the bottom of the plane?

it's 5 degrees

Answer 1

h=21 m
M=4 kg
At the top the block is at rest.
Total mechanical energy
= Kinetic energy + Potential energy
=0 + Mgh

At the bottom, the height of the block is zero
Total energy
=Kinetic energy at bottom + P. E. at bottom
= K + 0
=(1/2) M v^2
where , v is the velocity along the plane.
Hence,
v^2 = 2 g h
v=sqrt(2 h g)

I hope you can solve it now.

NOTE: The component of the velocity in the vertical and the horizontal directions will of course depend on the angle of inclination. But not the speed of the block as it is determined by the law of conservation of energy.

Cheers.

Answer 2

first thing's first, draw a FBD. Hopefully you already know that.

after drawing a FBD, you'll see that Fweight(x) of the block is the Fnet because there is no Friction involved. Because velocity changes, there is an acceleration. Therefore:

a = Fweight(x) / m

a = sin(5)mg / m

the mass cancel out, you left with:
a = sin(5)g
a = sin(5)9.8
a = .854m/s^2

because the block is traveling on the hypotenue of the incline, not the height, so we have to find the hypotenue.

H = 21 / sin(5)
H = 240.9m

Vf^2 = 2ad + Vi^2
Vf^2 = 2(.854)(240.9) + 0^2
Vf^2 = 411.45
Vf = 20.28m/s

hope this helps.

Answer 3

ok.E = 339 KeV = 339*10^3*a million.6*10^-19 J all of us understand capability = h(planck's consistent) * frequency h=6.626*10^-34 Js hence frequency(f) = (339*10^3*a million.6*10^-19)/(6.626*10^-34) From this answer c = Lambda(wavenlength) * f the place c = 3*10^8 m/s hence lambda = c/f

Answer 4

well technically when the block is at the bottom of the ramp wouldnt it stop??umm yeah so im gonna say zippo..0

Answer 5

Here is a web page with a very nice derivation for you!
http://www.glenbrook.k12.il.us/gbssci/Phys/Class/vectors/u3l3e.html