A 2.66 kg block is pushed along a horizontal floor by force p=138N directed at a

Question

A 2.66 kg block is pushed along a horizontal floor by force p=138N directed at an angle theta=24.8 degrees below the horizontal. The coefficient of kinetic between the block and the floor is 0.25. Calculate a) the magnitude of frictional force on the block and b) the acceleration of the block.

A 124 lb student on a steadily rotating ferris wheel has an apparent weight of 50.16lb at the highest point, a) what is the student's apparent weight a the lowest point? b) what is the student's apparent weight at the highest point, if the wheel's speed is doubled? Plse help thanks

Explanation / Answer

Mass of block, m = 2.66 kg Force on block, F = 138 N The angle of the applied force, = 24.80 The coefficient of kinetic friction, k = 0.25 a) The frictional force, fk = kmg = 0.25 * 2.66 * 9.8 = 6.52 N b) The net force, ma = Fcos - fk a = (Fcos - fk)/m = (125.3 - 6.52)/2.66 = 44.6 m/s^2 PLEASE GIVE THE NEXT QUESTION AS ANOTHER POST, I WILL SOLVE IT.                                                             THANq

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