- #1

aeroengphys

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## Homework Statement

"A block of mass 3.5kg slides down an inclined plane of length 6m that makes an angle of 60° with the horizontal. The coefficient of kinetic friction between the block and the incline is 0.3. If the block is released from rest at the top of the incline, what is its speed at the bottom. Assume that the acceleration due to gravity is 10m/s².

## Homework Equations

PE(top) = KE (bottom) + W(nc)

F=ma

## The Attempt at a Solution

mgh = 1/2mv² + Fscosθ

(3.5kg)(10m/s²)(5.2m) = 1/2(3.5kg)(v²) + (F)(6m)(cos60)

...F = ma

...F|| - Ff = ma

...mgsinθ - μmgcosθ = ma

...gsinθ - μgcosθ = a

...(10m/s²)(sin60) - (.3)(10m/s²)(cos60) = a

...7.16m/s² = a

182J = 1.75(v²) + (3.5kg)(7.16m/s²)(6m)(cos60)

v² = 61.04m/s

**v = 7.8 m/s**

The one thing I wasn't sure of was whether or not I substituted for F correctly in the Fscosθ part. If you could let me know if I did that correctly, I'd appreciate it. Thanks in advance.