a 2-kg-block is pushed upward from underneath by a constant 50-N vertical force.

Question

a 2-kg-block is pushed upward from underneath by a constant 50-N vertical force. The block is sliding on the bottom side of an inclined plane that makes an angle of 70 degrees with the vertical.

A 2-kg-block is pushed upward from underneath by a constant 50-N vertical force. The block is sliding on the bottom side of an inclined plane that makes an angle of 70 with the vertical. The coefficient of kinetic friction between the block and the plane is 0.2. If after traveling a total distance of 4 m the block is traveling at a speed of 4.443 m/s up the incline, what was the initial speed of the block 4.

Explanation / Answer

[-N*sin(theta) - fk]*S = change in KE = 0.5*m*(v^2-u^2) = 0.5*2*(4.443^2-u^2)

N*cos(theta) +F = m*g

N*cos(70)+50 = 2*9.81 = 19.62

N = -88.82 N

then (88.82*sin(70) + 0.2*88.82 )*4 = 0.5*2*(4.443^2-u^2)

405 = (4.443^2-u^2)

v= 19.62 m/s

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