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Chapter 5: Problem 20

A block slides down an inclined plane of slope angle \(\theta\) with constant velocity. It is then projected up the same plane with an initial speed \(v_{0} .(a)\) How far up the incline will it move before coming to rest? \((b)\) Will it slide down again?

Short Answer

Expert verified
The block will move a distance \(d = \frac{v_0^2}{2g \sin(\theta) } \) up the slope before coming to rest and it will not slide down again.

Step by step solution

01

Finding the coefficient of kinetic friction

Since block is sliding down the slope with a constant velocity, it implies that net force along the slope is zero. Let \( m \) be the mass of the block, then force due to gravity acting along the slope is \( mg \sin(\theta) \) and force due to friction is \( \mu_k mg \cos(\theta) \). Writing Newton's second law in the direction along the incline: \( \mu_k mg \cos(\theta) = mg \sin(\theta) \). From here we can solve for \( \mu_k \) which is the coefficient of kinetic friction: \( \mu_k = \tan(\theta) \).
02

Finding the acceleration when the block is projected up the slope

When it is projected upside, the net force acting on the block along the slope is \( F_{net} = mg \sin(\theta) - \mu_k mg \cos(\theta) = mg [ \sin(\theta) - \mu_k \cos(\theta) ] = mg [ \sin(\theta) - \tan(\theta) \cos(\theta) ] = 0 \). Here, we've substituted \( \mu_k = \tan(\theta) \) which we found in step 1. This gives the acceleration of the block when it is projected up the slope as \( a = F_{net}/m = 0 \)
03

Finding the distance it will move before coming to rest

Since acceleration of the block is zero, it will move with constant velocity \( v_0 \) and stops when frictional force balances the initial kinetic energy. Therefore, the distance \( d \) it will travel can be found using energy principle: \( \frac{1}{2} m v_0^2 = \mu_k m g cos(\theta) d \). Substituting \( \mu_k = \tan(\theta) \) and simplifying it, we get \( d = \frac{v_0^2}{2g \sin(\theta) } \)
04

Will it slide down again?

Since the net force acting on the block is zero, the block will not accelerate downwards once it has stopped. Therefore, it will not slide down the slope again.

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