Question

A crate slides down an incline plane and reaches the horizontal surface with a speed of 5 m/s. It slides for 2 m along this surface, with a coefficient of kinetic friction of 0.3. What is the velocity of the crate at the end of the 2 m?

Answer 1

\[v _{i} = 5 m/s\] The force slowing it down is kinetic friction, so \[F _{s} = \mu _{k} mg\] and the acceleration is \[F _{k}/m = \mu _{k} g\] and is negative (slowing). Now we can use kinematics\[v _{f}^{2} = v _{i}^{2} -2 \mu _{k}g \Delta x\] where the negative sign is due to the negative acceleration. solve for the final velocity

Answer 2

I think solving this problem using conservation of energy is the best way. ->\[(1/2)mv _{0}^{2}\] = \[\mu _{k}mg + (1/2)mv ^{2}\]
Since mass(m) can be divided in both sides of equation, you can easily get the answer(v) when you substitute numbers into each variables

Answer 3

You know the equation you gave can't be right because it is dimensionally incorrect... the units of the frictional force are Newtons, but the units of kinetic energy are Joules (Newton-meters). If you want to use this approach you have to calculate the work the frictional force does and set it equal to the change in kinetic energy...
\[- \mu _{k}mg \Delta x = 1/2m(v _{f}^{2} - v _{i}^{2})\]Note this is exactly the last equation given in my first response...

Answer 4

oops! my mistake. Thanks :)

Answer 5

Thank-you to both of you ;)