Question

A bob of mass of 0.18 kilograms is released from a height of
45 meters above the ground level. What is the value of the kinetic energy gained by the bob at the ground level?

a) 2.4 joules
b) 5.1 joules
c) 8.1 joules
d) 10.0 joules

Answer 1

hi ,

to calculate this quickly i would do the following:
remember th Ep = Potential Energy?

well your clue in your question is GAIN IN Ek(Kinetic energy), so all i would do is remember the following:

Lost in Ep = Gain in Ek

Answer 2

so calculate your Ep, and that will be your gain in Ek

Answer 3

remember energy cannot be created or destroyed, but can be changed from one form to the other. Potential energy can be changed to Kinetic energy.

hope this helps.

Ep = mgh = Gain in
\[E _{k} = \frac{ 1 }{ 2 }mv ^{2}\]

Answer 4

So would you just do \[E _{k} = \frac{ 1 }{ 2 }mv ^{2}\] to get the answer?

Answer 5

I think so, but you need to solve for the final velocity in order to solve for the Ek

Answer 6

How do you do that?

Answer 7

my knowledge regarding physics is maybe not the greatest but there is no need to solve velocity, in fact it would be difficult to solve without variables such as time and acceleration.
you do have to variables which is mass and distance.

consider a stationary object that possesses 81J of energy because of its HEIGHT of 45m above a reference plane. The 81J is potential energy. If the object is then allowed to fall freely (and the losses, such as the conversion of kinetic energy into heat energy are ignored), the potential energy will be nil and the kinetic energy 81J.

\[.^{.}.\] Lost in Ep = Gain in Ek = 81J

hope this makes sense

Answer 8

my mistake you can solve velocity, in fact quite easily.
consider the following:

m = .18kg; height = 45m; g = 10m/s
1. Lost in Ep
\[E _{p} = mgh\]
\[ E _{p} = .18kg(10m/s)45m\]
\[E _{p} = 81J\]

2. Now we can calculate velocity:
Remember the following?
\[E _{k}gained = E _{p}lost\]
so we can write it as
\[\frac{ 1 }{ 2 }mv ^{2} = mgh\]

so lets put in our variables
\[\frac{ 1 }{ 2 }(0.18kg)v ^{2} = 0.18kg(10m/s)45m\]
\[.^{.}.v ^{2} = \frac{ 0.18kg(10m/s)45m \times 2 }{ 0.18kg }\]
\[v ^{2} = 900\]
\[\sqrt{v ^{2}} = \sqrt{900}\]
\[v = 30m/s\]

3. Now to prove that the calculations above is correct:
\[E _{k } = \frac{ 1 }{ 2 }mv ^{2}\]
\[E _{k} = \frac{ 1 }{ 2 }(0.18kg)(30m/s)^{2}\]
\[E _{k} = 81J\]

4 . You can clearly see now that:
\[E _{p}Lost = E _{k}Gained\]

5. For exam purposes you would probably do it this way.

6.Check file send , only six pages long , it should help to understand kinetic energy better, pay attention to page 4.

Answer 9

@raffle_snaffle your answer is 81J, which look like option c, are you sure its 8.1J ?