hi guys
can someone plz help me with maj\june 2002 Q18. i m trying my level best but cant understand n do the question.
thanks in advance.
consider p.e at P as X. So p.e at Q is (X-50) J
k.e at P is 5kJ...that means p.e+k.e at P is X+5 kJ and W.d against friction is 10kJ. therefore at the lowest point it has (X+5)-10 kJ of energy=X-5 kJ
p.e at Q is X-50...the difference is....k.e at Q is given by
X-5-(X-50)= 45kJ
PS.i didnt solve this someone else did