Question

select the equation that shows that the gravitational potential energy of a 700 - kg boulder raised 8 m above ground level in units of n/kg, because m/s² is equivalent to n/kg.)
○ pe = \frac{mgh}{2}=\frac{(700\text{ kg})(10\text{ n/kg})(8\text{ m})}{2}=56000\text{ j}
○ pe = \frac{mg}{h}=\frac{(700\text{ kg})(10\text{ n/kg})}{(8\text{ m})}=56000\text{ j}
○ pe = mgh=(700\text{ kg})(10\text{ n/kg})(8\text{ m})=56000\text{ j}
○ pe = \sqrt{mgh}=\sqrt{(700\text{ kg})(10\text{ n/kg})(8\text{ m})}=56000\text{ j}

Explanation:

Step1: Recall gravitational - potential - energy formula

The formula for gravitational potential energy is $PE = mgh$, where $m$ is the mass, $g$ is the acceleration due to gravity, and $h$ is the height.

Step2: Substitute given values

Given $m = 700$ kg, $g=10$ N/kg, and $h = 8$ m. Substituting into $PE=mgh$, we get $PE=(700\text{ kg})(10\text{ N/kg})(8\text{ m})$.

Step3: Calculate the result

$(700)(10)(8)=56000$ J.

Answer:

PE = mgh = (700 kg)(10 N/kg)(8 m) = 56000 J