Question

A 5.80–kg block is set into motion up an inclined plane with an initial speed of vi = 8.20 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of θ = 30.0° to the horizontal.
(a) For this motion, determine the change in the block's kinetic energy.
(b) For this motion, determine the change in potential energy of the block–Earth system.
(c) Determine the friction force exerted on the block (assumed to be constant).
(d) What is the coefficient of kinetic friction?

Answer 1

kinetic energy change $\Delta$E = $\frac{5.8(8.2)^2}{2} = 195 J.$

potential energy change $\Delta$P = $5.8\cdot10\cdot3sin30 = 87J;$

friction force $F = \frac{195-87}{3} = 36 N;$

kinetic friction coefficient $\mu$ = $\frac{36}{58cos30} = 0.72$