As shown in the figure below, a box of mass m = 68.0 kg (initially at rest) is p

Question

As shown in the figure below, a box of mass

m = 68.0 kg

(initially at rest) is pushed a distance

d = 92.0 m

across a rough warehouse floor by an applied force of

FA = 214 N

directed at an angle of 30.0° below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.100. Determine the following. (For parts (a) through (d), give your answer to the nearest multiple of 10.)

(a) work done by the applied force

WA =   

(b) work done by the force of gravity

Wg =   J

(c) work done by the normal force

WN =  J

(d) work done by the force of friction

Wf =  J

(e) Calculate the net work on the box by finding the sum of all the works done by each individual force.

WNet =  J

(f) Now find the net work by first finding the net force on the box, then finding the work done by this net force.

WNet =  J

Explanation / Answer

Given that

Ass box of mass m = 68.0 kg

Distance pushed is d = 92.0 m

Applied force of FA = 214 N

The force directed at an angle of (theta) =30.0° below the horizontal.

The coefficient of kinetic friction between the floor and the box is (uk) = 0.100.

Now consider the force applied along the horizontal direction then

Facosthea-f =ma

(214N)cos30-fk =68kg*a-----(1)

Now the force acting in the vertical direction is

N =mg+Fasin30 =(68kg)(9.81)+(214N)(1/2) =774.08N

We can write fk =ukN =(0.100)(774.08) =77.408N

a)

The work done by the applied force is given by

W =Fa.d =Fadcostheta =Facos30*d =(214N)(92m)(0.866) =17,050J

Now the work done by force of gravity is

W =0J      as theta =90degrees

Now the work done by the normal force is

W =0J

And then the work done by force of friction is

W =-fk*d =-(77.408N)(92m) =-7121.536J

The net work on the box by finding the sum of all the works done by each individual force is

Wnet =17,050J-7121.536N =9928.464J

The net work by first finding the net force on the box, then finding the work done by this net force.

Fnet =Facos30-fk =(214N)cos30 -(77.408N) =107.921N

W =Fnet*d =(107.92N)(92m) =9928.772J

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