Question
A 1kg object has a potential energy of 20 J while stationary at the top of an incline. Neglecting friction, what is its speed halfway down the incline?
Answers
Henry
PE = 20J. at top of incline.
PE = 10J. halfway down the incline.
KE + PE = 20
KE + 10 = 20
KE = 10J.
KE = 0.5M*V^2 = 10
M = 1kg.
Solve for V.
PE = 10J. halfway down the incline.
KE + PE = 20
KE + 10 = 20
KE = 10J.
KE = 0.5M*V^2 = 10
M = 1kg.
Solve for V.