Part 1 A time t = 0, a 100kg box is at rest and is located at x = 0. The coeffic

Question

Part 1

A time t = 0, a 100kg box is at rest and is located at x = 0. The coefficient of static friction is 0.2. The coefficient of kinetic friction is 0.1 After time t = 0. A force of 150N is applied is the positive x direction. Where is the box after 20 seconds?

Part 2

A time t = 0, a 100kg box is very slowly moving and is located at x = 0. The coefficient of static friction is 0.2. The coefficient of kinetic friction is 0.1 After time t = 0. A force of 150N is applied is the positive x direction. Where is the box after 20 seconds?

Explanation / Answer

PART 1 -

since block is at rest static friction acts

since u(static) * m * g > 150 N

the block doesnt move

position of block after 20sec is x= 0

PART 2

since block is already slowly moving kinetic friction acts on it

equation of motion is given by

F-u(kinetic)*m*g= ma

a= 0.519 m/sec^2

S = 1/2 a *t*t

S=0.5a*t*t = 103.8m

Position of blok after 20 seconds is

X = 103.8 m ( after 20 seconds)

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