The stored potential energy of
E = (1/2)kX^2 = (1/2)(73.2)(0.167)^2
is first converted to kinetic energy. Then a decelerating frictional force
F = M g *0.148 is applied until FX' = E.
That is when it stops.
The distance that it slides over the rough surface is X' = E/F . Solve for that.
X' = [(1/2)(73.2)(0.167)^2]/[2.42*9.8*0.148 ]
A horizontal spring with spring constant 73.2 N/m is compressed 16.7 cm and used to launch a 2.42 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.148. How far does the box slide across the rough surface before stopping?
1 answer
1 answer