<span>364N should be your answer.. hope this helps
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In physics, power is defined as energy per unit time. You will also hear it described as work per unit time. The standard unit of measure for power is the watt, where a watt is defined as joules (energy) per second (time). This is expressed as a fraction as J/s. If you wanted to increase the power in any operation, you can either increase the energy (more joules) or reduce the time (fewer seconds).
A book falls to the floor.
A car skids to a stop.
A foam ball launches from a spring (Are the right answer, just did this one 4:34 pm Jan/21/19)
Answer:
v = sqrt[2*(F*h*cot(theta)-mgh)/m]
Explanation:
Work = KE + Ug
F*r=1/2mv^2+mgh
1/2mv^2=F*r-mgh
v=sqrt[2(F*r-mgh)/m]
r=h/tan(theta)=h*cot(theta)
