Tutorial Exercise A frictionless plane is 10.0 m long and inclined at 42.5°. A s

Question

Tutorial Exercise A frictionless plane is 10.0 m long and inclined at 42.5°. A sled starts at the bottom with an initial speed of 5.15 m/s up the incline. When the sled reaches the point at which it momentarily stops, a second sled is released from the top of the incline with an initial speed vi. Both sleds reach the bottom of the incline at the same moment. (a) Determine the distance that the first sled traveled up the incline. (b) Determine the initial speed of the second sled. Step 1 First we consider the free-body diagram below. This diagram applies to either sled on the frictionless slope of inclination = 42.5% ng The acceleration of the sled is directed down the incline, which we choose to be the positive x-direction. This acceleration has a magnitude of ,- mgsin gsins (a) If the first sled starts up the incline with speed Vox -5.15 m/s at the bottom, the distance it travels up the incline before stopping is m/s)21 2(9.80 m/s2 in m. Submit ikanoucannot corne back)

Explanation / Answer

A) By third equation of motion,

v^2 = u^2 + 2as

0 = 5.15^2 - 2*9.8*sin 42.5 degree *s

s = 5.15^2 /(2*9.8* sin 42.5 degree)

= 2 m answer

B) time taken by first sled to reach bottom = sqrt(2d/a)

= sqrt(2*2/(9.8* sin 42.5 degree))

= 0.777 s

Now time taken by second sled is also same, so by second equation of motion,

10 = u*0.777 + 0.5*9.8* sin 42.5 degree *0.777^2

10 - 2 = 0.777u

u = 8/0.777

= 10.3 m/s answer

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