Respuesta :
a. The potential energy and its kinetic energy at the instant it is dropped will be 1126.25 J and 0 J respectively.
b. The potential energy and its kinetic energy at the time it is 15m high will be 735.75 J and 390.25 J respectively.
c. The velocity when it is 15m high will be 156.1 m/sec.
d. The height when velocity is 12 m/s will be 15.62 m
e. The potential energy and its kinetic energy at the time it strikes the ground will be 0 J and 1126.25 J respectively
f. The velocity with which it hits the ground will be 21.22 m/sec.
What is kinetic energy?
The energy of the body by the virtue of its motion is known as the kinetic energy of the body. It is defined as the product of half of mass and square of the velocity.
According to the law of conservation of energy, the total energy is defined as the sum of kinetic energy and potential energy.
Total energy = kinetic energy+potential energy
Also, the energy can not be created nor be destroyed can be transferred from one form to another form.
Kinetic energy is due to the position.While the kinetic energy is due to the velocity.
a)The potential energy and its kinetic energy at the instant it is dropped.
[tex]\rm PE_{drop}=mgh \\\\\ PE_{drop}=5 \times 9.81 \times 25 \\\\ PE_{drop}=1126.25 \ J[/tex]
[tex]\rm KE_{drop}=\frac{1}{2} mv^2 \\\\ v= 0 \\\\ KE_{drop}=0 \ J[/tex]
TE= 1126.25+0
TE= 1126.25+0
b. Potential energy and its kinetic energy at the time it is 15m high.
[tex]\rm PE_{15}=mgh \\\\ \rm PE_{15}= 5 \times 9.81 \times 15 \\\\ PE_{15}=735.75 \ J[/tex]
TE=KE+PE
1126.25=KE+735.75
KE=390.25
c) Velocity when it is 15m high
[tex]\rm KE=390.25\\\\ 390.25=\frac{1}{2} \times 5 \times v_{15}^2 \\\\ v_{15}=156.1\ m/sec[/tex]
d. Height when velocity is 12 m/s.
[tex]\rm KE = \frac{1}{2}mv^2 \\\\\ KE =\frac{1}{2} \times 5 \times (12)^2 \\\\ KE=360\ J[/tex]
TE=KE+PE
1126.25=360+PE
PE=766.25
PE=766.25
mgh=766.25
5 ×9.81 ×h=766.25
h=15.62 m
E. Potential energy and its Kinetic energy at the time it strikes the ground.
[tex]\rm PE_{ground}=0 \ J[/tex]
TE=KE+PE
TE=KE+0
TE=KE
TE= 1126.25
KE=1126.25 J
f. Velocity with which it hits the ground.
[tex]\rm KE_{ground}=1126.25 J \\\\ \frac{1}{2} mv^2=1126.25\\\\ v^2 = 450 \\\\ v= 21.22 \ m/sec[/tex]
To learn more about kinetic energy refer to the link;
brainly.com/question/999862
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