♣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;
♠ kinetic energy of the pig at the moment it’s 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;
♦pig’s 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=μ*w1, w1=mg*cos(b) is component of pig’s weight normal to the aluminum ramp; or;
f= μ*mg*cos(b);
♦Thus E2=E-E1; or; f*L = 0.5*m*v^2 –mgh; or;
(μ*mg*cos(b)) * (h/sin(b)) =0.5*m* 2as –mgh; or;
μ*gh*cos(b)/sin(b) =as –gh, hence cot(b) =(as –gh)/(μ*gh), hence
(a); b=atan(μ*gh/(as-gh))= atan(0.07*9.8*0.5/(4.9*1.8 -9.8*0.5)) =5°;
♥ pig’s potential energy above the low end of the ramp is
W= mg*(s*sin(b) +h); pig’s ramp position is p=s +L = s + h/sin(b);
Pig’s 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) – μ*mg*cos(b)*(s + h/sin(b));
v^2 =2g*(s*sin(b) +h)*(1 –μ*cot(b)) =2.575;
(b); v =1.605 m/s;
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