β£the distance the pig was pushed up is

s=0.5*a*t^2, where s=1.8m, a=4.9m/s^2, t is time of push-up;

speed of the pig gained after s=1.8m is v=a*t; thus v^2=2as;

β kinetic energy of the pig at the moment itβs released after push-up is

E=0.5*m*v^2, which is enough for the pig to move by itself L distance more along the ramp, where L=h/sin(b), h=50cm=0.5m, b is angle in question;

β¦pigβs potential energy at vertex (when stopped) is E1=mgh;

the energy E2=E-E1 is gone on friction, that is E2=f*L is work of friction, where

f=ΞΌ*w1, w1=mg*cos(b) is component of pigβs weight normal to the aluminum ramp; or;

f= ΞΌ*mg*cos(b);

β¦Thus E2=E-E1; or; f*L = 0.5*m*v^2 βmgh; or;

(ΞΌ*mg*cos(b)) * (h/sin(b)) =0.5*m* 2as βmgh; or;

ΞΌ*gh*cos(b)/sin(b) =as βgh, hence cot(b) =(as βgh)/(ΞΌ*gh), hence

(a); b=atan(ΞΌ*gh/(as-gh))= atan(0.07*9.8*0.5/(4.9*1.8 -9.8*0.5)) =5Β°;

Β

β₯ pigβs potential energy above the low end of the ramp is

W= mg*(s*sin(b) +h); pigβs ramp position is p=s +L = s + h/sin(b);

Pigβs final kinetic energy is W1=W-W2, where W2=f*p is work of friction;

and W1=0.5*m*v^2; thus

0.5*m*v^2 = mg*(s*sin(b) +h) β ΞΌ*mg*cos(b)*(s + h/sin(b));

v^2 =2g*(s*sin(b) +h)*(1 βΞΌ*cot(b)) =2.575;

(b); v =1.605 m/s;

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