Modern Physics For Scientists And Engineers
Modern Physics For Scientists And Engineers
2nd Edition
ISBN: 9781938787751
Author: Taylor, John R. (john Robert), Zafiratos, Chris D., Dubson, Michael Andrew
Publisher: University Science Books,
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Chapter 1, Problem 1.1P
To determine

(a)

To calculate:

The block's position as a function of time for Oxy axes and the time taken by the block to reach the bottom.

Expert Solution
Check Mark

Answer to Problem 1.1P

Position of the block in terms of (x,y) = (( gsinθμgcosθ2)t2,0)

Time taken by the block to reach the ground t=2lgsinθμgcosθ

Explanation of Solution

Given:

A block of mass msliding down on an inclined plane making angle θ with respect to the horizontal surface.

Formula used:

Using equation of motion,

  s=ut+12at2

s = distance travelled by the block

u = initial velocity

t = time taken by the block

a = acceleration of the block

Calculation:

Modern Physics For Scientists And Engineers, Chapter 1, Problem 1.1P , additional homework tip  1

Modern Physics For Scientists And Engineers, Chapter 1, Problem 1.1P , additional homework tip  2

Forces on the block along x-axis,

Since the block is moving in x direction therefore, by the equation of motion.

  mgsinθfx=ma.......(1)

  mgsinθ is the component of mg along x-axis

  fx is the frictional force

m is the mass of the block

a is the acceleration of the block

by Newton's law of motion,

  N=mgcosθ

N = normal force

  mgcosθ = vertical component of mg along y-axis

force of friction is given by

  fx=μN.......(2)

Put the value of N in equation (2)

  fx=μmgcosθ

Put the value of fxin equation (1)

  mgsinθμmgcosθ=magsinθμgcosθ=a......(3)

By the equation of motion

  x=uxt+12at2

  ux = 0 = initial velocity of the block

  x=12at2

Put the value of a from equation (3)

  12(gsinθμgcosθ)t2.......(4)

Since there is no motion in vertical direction,

Therefore, y =0

Time taken by the block to reach the ground.

Here, distance covered by the block is l.

Therefore, put x = l, in equation (4)

  l=12(gsinθμgcosθ)t2t= 2l ( gsinθμgcosθ )

Conclusion:

Therefore, position of the block can be defined in (x,y) as (( gsinθμgcosθ2)t2,0)

Time taken by the block to travel distance l, t=2l( gsinθμgcosθ)

To determine

(b)

To calculate:

The block's position as a function of time for Ox'y' axes and the time taken by the block to reach the bottom.

Expert Solution
Check Mark

Answer to Problem 1.1P

Position of the block in terms of (x,y) =(12acosθt2,12asinθt2)

Time taken by the block to reach the ground, t=2lgsinθμgcosθ

Explanation of Solution

Given:

A block of mass m sliding down on an inclined plane making angle θ with respect to the horizontal surface.

Formula used:

Using equation of motion,

  s=ut+12at2

s = distance travelled by the block

u = initial velocity

t = time taken by the block

a = acceleration of the block

Calculation:

Modern Physics For Scientists And Engineers, Chapter 1, Problem 1.1P , additional homework tip  3

Distance travelled in x-direction

  x=lcosθ=uxt+12axt2where, axis the component of acceleration in x directionax=acosθux= initial velocity = 0x=lcosθ=12acosθt2.......(1)

Distance travelled in y direction

  y=lsinθ=uyt+12ayt2where, axis the component of acceleration in x directionay=asinθuy= initial velocity = 0y=lsinθ=12asinθt2.......(2)

Position of the block in (x,y) =(12acosθt2,12asinθt2)

Time taken by the block to reach the ground

Adding equation (1) and (2)

  lcosθlsinθ=12acosθt212asinθt22l(cosθsinθ)=at2(cosθsinθ)2l=at2t= 2laput a=gsinθμgcosθ from equation (3) of part (a)t= 2l gsinθμgcosθ

It is clear from the solution of part (a) and (b), that time taken by the box is same to travel the distance l.

Taking Oxy axes is less convenient because it is difficult to draw free body diagram for the object and so the forces and their directions.

Conclusion:

For Oxy axes, time taken by the block to reach the ground t=2lgsinθμgcosθ

For Oxy axes, position of the block in (x,y) is given as (12acosθt2,12asinθt2)

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