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Free_Kalibri [48]
4 years ago
7

How much work w is done by the electrostatic force on the moving point charge?

Physics
1 answer:
V125BC [204]4 years ago
3 0

As we know that electrostatic force is a conservative force

so we can say by the condition of conservative force

F_c = -\frac{dU}{dr}

here we can rearrange the above equation as

dU = - F_c.dr

now integrate both sides

\int dU = - \int F_c . dr

Now we know by the definition of work done by a force is given by

W = \int F.dr

now work done by conservative force is given as

W_c = \int F_c . dr

Now from above work done by electric field to move charge from one point to other is given as

W_{EF} = \int F_{e}.dr = - \int dU

so here work done is given as

W_{EF} = U_i - U_f

so change in potential energy is given by work done

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When is a zero not significant?
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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
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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
a lorry travels 3600m on a test track accelerating constantly at 3m/s squared from standstill. what is the final velocity (3 sig
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The final velocity of the truck is found as 146.969 m/s.

Explanation:

As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.

Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.

2as =v^{2} - u^{2} \\\\v^{2} = 2as + u^{2}

Then, the final velocity will be

v^{2} = 2 * 3 * 3600 + 0 = 21600\\ \\v=\sqrt{21600}=146.969 m/s

Thus,  the final velocity of the truck is found as 146.969 m/s.

8 0
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