A block is pushed up an incline by an applied force that is directed parallel to

Question

A block is pushed up an incline by an applied force that is directed parallel to the incline. The coefficient of kinetic friction between the block and the incline is (0.2). The mass of block is 5 Kg, and the incline angle is 30 degrees. Also, the block moves up the incline at a constant speed of 3.5 m/s. After the block has moved 1.5 meters up the incline, a) Find the work done by gravity b) Find the work done by the applied force c) Find the work done by friction d) Find the change in kinetic energy of the block

Explanation / Answer

a)work done by gravity = weight*vertical displacemnt=-5*9.8* 1.5*sin 30 =-36.75J

b)work done by applied force = force*displacement

now applied forec =force of friction + weight*sin 30 = 0.2*W*cos30+W*sin30=0.2*5*9.8*sin 30 + 5*9.8*cos 30=47.33N

So work=47.33*1.5=71J

c)Work done by friction = force of friction*displacemnt=0.2*W*cos30*1.5=0.2*5*9.8*sin 30*1.5=-14.52J

d)change in kinectic energy = 0 because velocity does not change.

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