A 4.3 kg block is projected at 5.0 m/s up a plane that is inclinedat 30 o with t

Question

A 4.3 kg block is projected at 5.0 m/s up a plane that is inclinedat 30o with the horizontal.

(a.) How far up along the plane does the block go if the plane isfrictionless?
(b.) How far up along the plane does the block go if thecoefficient of kinetic friction between the block and the plane is0.40?
(c.) In the latter case, what is the increase in thermal energy ofblock and plane during the block's ascent?
(d.) If the block then slides back down against the frictionalforce, what is the block's speed when it reaches the originalprojection point?

Explanation / Answer

(a) the distance traveled up: mg*d*sin30=m*v2/2 => d=2.55m (b)the kinetic energy transfers to the potential energy andwork done by the friction force: m*v2/2 =m*g*cos30*0.4*d +mg*d*sin30 => d=1.51m (c) increase in thermal energy of block and plane duringthe block's ascent= work done by friction force m*g*cos30*0.4*d=Q=22J (d) we have : m*g*1.51*sin30 =m*g*cos30*0.4*d +m*v2/2 v=2.14 m/s (d) we have : m*g*1.51*sin30 =m*g*cos30*0.4*d +m*v2/2 v=2.14 m/s

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