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3 years ago

If an object has 100 J of PE, how much work can it do (with that PE)?

Physics

1 answer:

Stella [2.4K]3 years ago

4 0

Answer:

W = 100 J

Explanation:

The potential energy of an object is given by :

$P=mgh$

It is said to possessed by an object due to its position under the action of gravity.

The work done by an object is given by :

W = F×d , F is force and d is displacement

We know that, F = mg

So, if an object has 100 J of PE, it means that it can do the work of 100 J.

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Refer to Concept Simulation 4.4 for background relating to this problem. The drawing shows a large cube (mass = 28.9 kg) being a

Usimov [2.4K]

Answer:

smallest magnitud is P=33.3 N

Explanation:

We are analyze the situation as an external force is applied and there is a friction force. We have a problem with Newton's second law.

F = ma

As the two blocks go together they must have the same acceleration, so we can calculate this for the entire system

P = (m1 + m2) a

a = P / (m1 + m2)

In this case there is no friction force because the small block does not touch the ground.

In order to calculate the friction force, we must analyze each system component separately.

The large block on the X axis has an applied P force and as it moves feels a force from the small block. In the Y axis has the weight (W1) and the reaction to normal (N1)

For the small block on the X axis, the force it feels is the thrust of the large block, note that this is an action and reaction force between the two blocks, it is the same definition we have of the normal one, so we can call this force (N)

Y axis it has the weight (W2) down, the force of friction (fr) that opposes the movement, so it is directed upwards. we write these equations

N = m2 a

fr -W2 = 0

fr = W2

The definition of friction force is

fr = μ N

Let's replace and calculate

μ (m2 a) = m2 g

μ (P / (m1 + m2)) = g

P = g /μ (m1 + m2)

Let's calculate the value of this force

P = 9.8 / 0.710 (28.9 +4.4)

P = 13.80 (33.3)

P = 33.3 N

This is the minimum friction force that prevents the block from sliding down

6 0

3 years ago

Bart stole a watermelon and ran 5,000 feet from the cops and they chase lasted 0.1 hours how fast was Bart running in miles per

liraira [26]

I am not as sure but I think it is 9.469 miles

5 0

3 years ago

Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan

Eddi Din [679]

To increase the centripetal acceleration to $2.00 m/s^2$ , you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

$a=0.50 m/s^2$

We want to increase it by a factor of 4, i.e. to

$a'=2.00 m/s^2$

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

This way, the new acceleration is

$a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a$