a 600kg box is being pushed across the floor by a machine that constantly applies 400N of force.  Considering a frictional force of 75N between the box and the floor, what is the resulting...

a 600kg box is being pushed across the floor by a machine that constantly applies 400N of force.  Considering a frictional force of 75N between the box and the floor, what is the resulting acceleration of the box  What is the weight of the box.  How much force would the machine need to provide in ordder for the box to accelerate at 4.2m/s^2?

Asked on by mpumpkin

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aruv | High School Teacher | (Level 2) Valedictorian

Posted on

Let us assume floor is parallel to horizontal. Thus normal force is perpendicular to the floor. Thus equation of frictional force is

`F=muN` ,  `mu` and N are respectively coefficient of friction and normal force.

Mass of the box=600 kg

Weight of the box= 600 g N , g is acceleration due to gravity.

But weight is equal to normal force.

Given frictional force =75 N

Thus

`mu=F/N=F/W=75/600 =.125`

Box is pushed by  a force= 400N

Frictional force opposes motion i.e its direction is opposite to force.

Thus resultant force acting on box which cause motion=400-75

=325 N

By Newton's second law of the motion

`F_r=ma`

`325=600a`

`a=325/600`  m/sec^2

(ii)The required acceleration =4.2 m/sec^2

 Let requires force be `F_n` then by Newton's second law of motion

`(F_n-75)=4.2xx600 `

`F_n=2520+75=2595N`

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