Answer :
Answer:
[tex]534\; {\rm J}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
Explanation:
The mechanical energy of an object is the sum of the potential energy and kinetic energy of that object. During a free fall where air resistance is negligible, the reduction in the potential energy of the object should be equal to the increase in the kinetic energy of the object.
For an object of mass [tex]m[/tex], if the change in height is [tex]\Delta h[/tex] and gravitational field strength is constantly [tex]g[/tex], the change in the gravitational potential energy of the object would be:
[tex](\text{change in GPE}) = m\,g \, \Delta h[/tex].
In this question, the mass of the object is [tex]m = 3.0\; {\rm kg}[/tex], and the change in height is [tex]\Delta h = (-50)\; {\rm m}[/tex]. Note that the change in height is negative because the object would be below the initial position.
The change in the gravitational potential energy of this object would be:
[tex]\begin{aligned} & (\text{change in GPE}) \\ =\; & m\,g \, \Delta h \\ =\; & (3.0\; {\rm kg})\, (9.81\; {\rm m\cdot s^{-2}})\, ((-50)\; {\rm m}) \\ =\; & (-1471.5)\; {\rm J}\end{aligned}[/tex].
In other words, the gravitational potential energy of this object would be [tex]1471.5\; {\rm J}[/tex] lower than the initial value.
For an object of mass [tex]m[/tex], if velocity changes from an initial value of [tex]u[/tex] to a new value of [tex]v[/tex], the change in the kinetic energy of the object would be:
[tex]\begin{aligned}(\text{change in KE}) &= \frac{1}{2}\, (m)\, \left(v^{2} - u^{2}\right)\end{aligned}[/tex].
In this question, the velocity of this [tex]m = 3.0\; {\rm kg}[/tex] has changed from [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] to [tex]v = (-25)\; {\rm m\cdot s^{-1}}[/tex] (negative because the object is moving downwards.) The change in the kinetic energy of this object would be:
[tex]\begin{aligned} & (\text{change in KE}) \\ =\; & \frac{1}{2}\, (m)\, \left(v^{2} - u^{2}\right) \\ =\; & \frac{1}{2}\, (3.0\; {\rm kg})\, \left((25\; {\rm m\cdot s^{-1}})^{2} - (0\; {\rm m\cdot s^{-1}})^{2}\right) \\ =\; & 937.5\; {\rm J}\end{aligned}[/tex].
In other words, the object has gained [tex]937.5\; {\rm J}[/tex] of kinetic energy.
In a free fall where the effect of air resistance is negligible, the increase in the kinetic energy of the object should be equal to the reduction in the gravitational potential energy. However, in this example, the increase in kinetic energy is [tex](1471.5\; {\rm J} - 937.5\; {\rm J}) = 534\; {\rm J}[/tex] less than the reduction in gravitational potential energy. The difference is likely the result of the air resistance on this object.
