Physics Motion on an incline

Question

A block starts sliding up an incline with an initial velocity of v₀ = 4 m/s, as shown.

If the angle of the inclined plane, � = 15 degrees, and the coefficient of kinetic friction between the block and the plane is 0.2, what distance from it's starting position does the block travel along the plane before it comes to a stop?

Answer 1

gravitational deceleration = –g sinθ

frictional deceleration = –µg cosθ

both decrease velocity going up the incline

total negative acceleration = a = –g(sinθ + µcosθ) =

–9.8(sin15° + 0.2cos15°) ≈

–9.8(0.452) ≈ –4.4296

From kinematics we have

v² – v₀² = 2ax

where x is the distance the block travels along the plane before it comes to a stop.

Because v = 0 when the block comes to a stop, this becomes

–v₀² = 2ax

Solving for x gives

x = –v₀²/2a

Plugging in a = –4.4296 and v = 4 gives

x = –v₀²/2a = –(4)²/(2a) = –16/(2a) = –8/a =

8/4.4296 = 1.806

Answer (arbitrarily rounded; maybe it should actually be just 1 significant figure)

The block travels 1.8 m up the incline.