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A block of mass m is pushed up a rough incline by a constant forceacting parallel to the incline. The block is displaced by d up theincline. Calculate; a.) the work done by gravity for thisdisplacement( not numerical). b.) the work done by the appliedforce( not numerical). c.) work done by kinetic friction usingkinetic friction kinetic( not numerical). d.) the net work done onthe block if F= 150 N, d= 1m, angle= 25 degrees, coefficientfriction( kinetic)= .3

Answer :

opudodennis

Answer:

(a)-4.15m  

(b) 150 Nm

(c) -2.67m

(d)150-6.813149m

Explanation:

(a)

The work done by the gravity is [tex]W = -mgdsin\theta[/tex] where m is mass, g is gravitational constant, [tex]\theta[/tex] is angle of inclination, F is force on the inclined plane and d is the displacement of the body in the plane.

W=-(m*9.81*1*sin 25)= -4.1458851m  

w--4.15m

Note that m here is mass, not units

(b)

The work done by the applied force is Wa = F *d=150*1=150 Nm

(c)

The work done by the frictional force is [tex]Wf = -d*(\mu *Normal force) [/tex]  

But the normal force is [tex]mgcos\theta[/tex]

[tex]Wf=-(\mu *mg cos\theta )*d[/tex]  =-(0.3m*9.81*cos25)= -2.6672638m

Wf=-2.67m

Where m is not units but mass

(d)

The net force is [tex]Fnet = F – mgsin\theta- \mu *mgcos\theta[/tex]

The work done by the net force is [tex]W = Fnet*d  =( F – mgsin\theta- \mu *mg cos\theta )*d[/tex]

W=(150-4.1458851m-2.6672638m)=150-6.813149m

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