A 500 N block is resting on an inclined plane and is subjected to a constant force of 600N acting parallel to the inclined plane. After the block has moved 3 m from rest along the inclined plane the force 600N is removed. The inclined plane has a slope of three vertical to four horizontal. Coefficient of friction is 0.20
a. Which is the following gives the distance that the black will move further along the inclined plane until it will stop?
b. Which of the following gives the velocity of the block when this for 600 Newton was removed?
c. Which of the following gives the velocity of the block when it returns to its initial position?
Extracted text: 500N 500 N V2=0 500N Vj=Qd 600 N B' 4 3m