Asked by Rosie
Doyle goes bowling with a cement brick to improve his score. If the brick has a mass of 1.5kg and an initial speed of 5.0m/s upon release a)how much kinetic energy did it possess upon release i.e. when let it go and b)how far will the brick slide before coming to a halt if the coefficient of friction is 0.25?
what I have found so far is the KE
a)19J
but b) I have no clue where to start please help
what I have found so far is the KE
a)19J
but b) I have no clue where to start please help
Answers
Answered by
Henry
b. Fk = u*Mg = 0.25 * 1.5*9.8 = 3.675 N.
PE = KE - Fk*d = 19-3.675*d = 1.5*9.8*d, 19-3.675d = 14.7d, -18.375d = -19, d = 1.03 m.
PE = KE - Fk*d = 19-3.675*d = 1.5*9.8*d, 19-3.675d = 14.7d, -18.375d = -19, d = 1.03 m.
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