Question

a 3.18-kg rock is released from rest at a height of 26.6 m. ignore air resistance and determine (a) the kinetic energy at 26.6 m, (b) the gravitational potential energy at 26.6 m, (c) the total mechanical energy at 26.6 m, (d) the kinetic energy at 0 m, (e) the gravitational potential energy at 0 m, and (f) the total mechanical energy at 0

Answer 1

Answer:

(a) 0 J

(b) 828.96 J

(c) 828.96 J

(d) 828.96 J

(e) 0 J

(f) 828.96 J

Explanation:

Given:

Mass of the rock is, $m=3.18\ kg$

Initial height of the rock is, $h_{i}=26.6\ m$

Final height of the rock is, $h_f=0\ m$

Initial velocity of the rock is, $v_i=0\ m/s(Rest)$

Acceleration due to gravity is, $g=9.8\ m/s^2$

(a)

Initial kinetic energy is given as:

$K_i=\frac{1}{2}mv_i^2$

Plug in the given values and solve for '$K_i$'. This gives,

$K_i=\frac{1}{2}\times 3.18\times 0^2=0$

Therefore, initial kinetic energy is 0 J.

(b)

Initial potential energy is given as:

$U_i=mgh_i\\U_i=3.18\times 9.8\times 26.6=828.96\ J$

Therefore, initial potential energy is 828.96 J.

(c)

Mechanical energy is equal to the sum of kinetic and potential energy. It is always a constant value. Therefore,

$ME=K_i+P_i\\ME=0+828.96=828.96\ J$

Therefore, mechanical energy at 26.6 m is 828.96 J.

(d)

Now, at the final height of 0 m, the decrease in potential energy will be equal to the increase in the kinetic energy. In other words, the potential energy at the start will be converted to kinetic energy at the bottom.

Therefore, kinetic energy at the 0 m is given as:

$K_f=U_i=828.96\ J$

(e)

Potential energy at the final height is given as:

$P_f=mgh_f=3.18\times 9.8\times 0=0\ J$

Therefore, final potential energy is 0 J.

(f)

Total mechanical energy is a constant and doesn't depend on the height.

So, total mechanical energy at 0 m is same as that at 26.6 m.

Total mechanical energy is 828.96 J.