posted by alex on .
As part of a training schedule a weightlifter pushes a heavy metal block
along the floor. The lifter applies a force of 850 N downwards on the block
at an angle of 65° to the vertical (25° to the horizontal). The weightlifter
pushes the block forwards 4 m in 1.6 s. Calculate the power output of the
weightlifter when moving the block 4 m forwards. [10 marks]
F = 850N @ 25 Deg.
Fx = 850*cos25 = 770.4 N. = hor. Force.
Fy = 850*sin25 = 359.2 N. = Ver. force.
P = Fx*V = Fx * (d/t).
P=770.4 * (4/1.6) = 1926 J/s=1926 J/s =