/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 63 A very slippery block of ice sli... [FREE SOLUTION] | 91影视

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Chapter 2: Q. 63 (page 62)

A very slippery block of ice slides down a smooth ramp tilted at angle u. The ice is released from rest at vertical height h above the bottom of the ramp. Find an expression for the speed of the ice at the bottom

Short Answer

Expert verified

The final velocity of the ice block is 2gh

Step by step solution

01

Step 1.  write the given information

The angle of inclination is u

The block of ice is initially at rest, therefore, vi=0m/s
The height at which the block was at rest is h

Let the final speed of the block isvf

02

Step 2. To determine the speed of the ice block

Let the slant height of the ramp is S

The acceleration experienced by the block is a=gsinu
here, gis the acceleration due to gravity

sin(u)=hS

Hence, the acceleration is expressed asa=ghS
Now, write the equation of motion,
vf2=vi2+2aS
Substitute the known values,
vf2=0+2ghSSvf2=2ghvf=2gh
Therefore, the final velocity of the ice block is2gh

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