A 1.0 kg wood block is launched up awooden ramp that is inclined at a 34°angle.

Question

A 1.0 kg wood block is launched up awooden ramp that is inclined at a 34°angle. The block's initial speed is 13m/s. (Use µk = 0.20 for the coefficient of kineticfriction for wood on wood.) (a) What vertical height does the block reachabove its starting point?
1 m
(b) What speed does it have when it slides back down to itsstarting point?
2 m/s down theramp

Please distinguish which answer is a and which is b.
(a) What vertical height does the block reachabove its starting point?
1 m
(b) What speed does it have when it slides back down to itsstarting point?
2 m/s down theramp

Please distinguish which answer is a and which is b.

Explanation / Answer

mass m = 1kg angle = 34 degrees initial speed u = 13 m / s coefficien tof kinetic friction = 0.2 (a). final velocity v = 0 distance travelled along the ramp S = [ v ^ 2 - u ^ 2 ]/ 2a where a = accleration = - [ g sin + gcos ]                                   =- [ 5.48 +1.624 ]                                 = -7.104 m / s ^ 2 So, S = 11.894 m Therefore vertical height   h = S sin                                         = 11.894 * sin 34 = 6.651 m (b). In down ward direction : ---------------------------- Initial velocity U = 0 Accleration a ' = g sin - g cos                        = 5.48 -1.624 = 3.856 m / s ^ 2 distance S = 11.894 m from the relation V ^ 2 -U ^ 2 = 2a' S speed when it slides back down to its starting pointis V = [ 2a' S ]    Since U =0                                                                                     = 9.577 m / s

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