A 4.3 kg block is projected at 4.8 m/s up a plane that is inclined at 30° with t

Question

A 4.3 kg block is projected at 4.8 m/s up a plane that is inclined at 30° with the horizontal.
(a) How far up along the plane does the block go if the plane is frictionless?
2.35 m THIS IS CORRECT

(b) How far up along the plane does the block go if the coefficient of kinetic friction between the block and the plane is 0.40?
m

(c) In the latter case, what is the increase in thermal energy of block and plane during the block's ascent?
J

(d) If the block then slides back down against the frictional force, what is the block's speed when it reaches the original projection point?
m/s

Explanation / Answer

a) a = gsin

a= 9.8sin30 = 4.9 m/s^2

using kienmatic equation,

v^2 - u^2 = 2as

0 - 4.8^2 = 2(-4.9)(s)

s= 2.35 m apprx

b) a(retardation) = gsin + gcos

a= 4.9 + 0.40(9.8)cos30 = 8.29 m/s^2 apprx

v^2 - u^2 = 2as

0 - 4.8^2 = 2(-8.29)(s)

s= 1.39 m apprx

c) increase in thermal energy = due to work done by friction

Incraese in tehrmal energy = forc eof frictiuon x dispalcemnet

= mgcos( 1.39) = 0.4 (4.3 )(9.8)cos30 ( 1.39) = 20.29 J apprx

d) PE at the slope of 1.39 m

PE = mg( 1.39 sin30)= mg( 0.6895) J = 4.3(9.8) (0.6895)=29.055 j apprx

KE at bottom= 29.055 - 20.29 = 8.76553 J apprx

0.5mv^2 = 8.76553

v = 2.02 m/s apprx

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