roadjester
  • roadjester
A block slides down a frictionless plane having an inclination of 15.0 degrees. The block starts from rest at the top, and the length of the incline is 2.00 m. a) Draw a free body diagram for the block. b) Find the acceleration of the block and c) its speed when it reaches the bottom of the incline.
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
would the force be mg sin(t) ? just curious
amistre64
  • amistre64
which i spose might take gravity from stright down to down the incline
amistre64
  • amistre64
|dw:1332700324332:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
i would interp this as; instead of having accel due to gravity being 9.8; that the incline reduces it to 9.8sin(15degrees) and since velocity is the integration of accel; v(t) = 9.8sin(15) t; and position from there is another integration to determine how far it moves in time ....
amistre64
  • amistre64
assuming im right :) a(t)=-9.8sin(15) v(t) = -9.8sin(15) t p(t) = -4.9sin(15) t^2 + 2sin(15) -4.9sin(15) t^2 + 2sin(15) = 0; when t = 2sqrt(5)/7 v(2sqrt(5)/7) = -9.8sin(15) * 2sqrt(5)/7 = abt -1.62 that is direction and speed; so speed = 1.62 if i did it right
anonymous
  • anonymous
@amistre64 \(g \sin(\theta)\) is correct for the magnitude along the incline. However, note that the length of the incline is 2 meters, therefore displacement is 2\[d = 2 \rightarrow 2 = {1 \over 2} 9.8 \sin(15) t^2\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.