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Chapter 6: Q. 48 (page 156)

An object of mass m is at rest at the top of a smooth slope of height hand lengthL. The coefficient of kinetic friction between the object and the surface, μk, is small enough that the object will slide down the slope after being given a very small push to get it started. Find an expression for the object’s speed at the bottom of the slope.

Short Answer

Expert verified

The velocity of an object down the slope isvf=2L(μkg-gsinθ)

Step by step solution

01

Step 1. Given Information

A smooth slope of height hand length L

The coefficient of kinetic friction isμk

02

Step 2. Find the speed of object

We need to determine the speed of an object at the bottom of the slope, it was at rest the top and started from here

From the diagram below,

Using Newton's second law, resolving all forces acting along xaxis, we get

F→net=f→k-F→Gsinθ

where θ=tan-1hL;fk⇀=μkmg;FG⇀=mg;F⇀net=ma

⇒ma=μkmg-mgsinθa=μkg-gsinθ

with this acceleration, the object will slide downwards hence to find speed we use the equation of motion,

vf2=vi2+2asvf2=02+2(μkg-gsinθ)×Lvf=2L(μkg-gsinθ)

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