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3 years ago

If a body of mass 5og moves in a circular path of radius 10cm .find work done

Physics

1 answer:

Sindrei [870]3 years ago

3 0

Answer:

0Nm, no work is done.

Explanation:

Work done is defined as the Force per distance meaning force times the distance moved in the direction of the force.

Now the body of mass 50g has a centripetal force acting on it directed towards the centre. Now in actuality the body stays along the circle it doesn't really move to the centre of the circle.

Hence the force doesn't move a distance, and so from the definition of work done;

F×d ; d =0

Hence work done = mv2/r × 0= 0Nm

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3 years ago

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2 years ago

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As a freely falling object picks up downward speed, what happens to the power supplied by the gravitational force?

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We will start from the definition of power in terms of the Force. Power could be described as the change of energy in an instant of time. Considering that Energy is the product between the Force and the distance traveled we would arrive at the expression

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3 years ago

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denis-greek [22]

1) Right answer: $9.8m/s^{2}$

Gravity is a physical force with which the Earth exerts on all bodies towards its center due to the fact its mass is greater than the bodies on its surface.

Nevertheless, it is important to note that this value is not the same in all the points of the planet, that is why an average value is used, which is $9.8m/s^{2}$ .

2) Right answer: $8m/s$

The Average Velocity Formula is:

$V_{average}=\frac{V_{i}+V_{f}}{2}$

Where $V_{i}$ is the Initial Velocity and $V_{f}$ is the Final Velocity.

Knowing this, let's apply the formula:

$V_{average}=\frac{0m/s+16m/s}{2}$

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$V_{average}=\frac{10m/s+28m/s}{2}$

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2 years ago

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A block with a mass of 31.8 kg is pushed on a frictionless

OlgaM077 [116]

Answer: Total work done on the block is 3670.5 Joules.

Step by step:

Work done:

$W = F.d.\cos\theta$

With F the force, d the displacement, and theta the angle of action (which is 0 since the block is pushed along the direction of displacement, and cos 0 = 1)

$W = F.d$

Given:

F = 75 N

m = 31.8 kg

Final velocity $v_f = 15.3 \frac{m}{s}$

In order to calculate the Work we need to determine the displacement, or distance the block travels. We can use the information about F and m to first figure out the acceleration:

$F = ma\\\implies a=\frac{F}{m}=\frac{75 N }{31.8 kg}\approx 2.36 \frac{m}{s^2}\\$

Now we can determine the displacement from the following formula:

$d = \frac{1}{2}a^2+v_0t+d_0$

Here, the initial displacement is 0 and initial velocity is also 0 (at rest):

$d = \frac{1}{2}at^2\\$

Now we still have "t" as unknown. But we are given one more bit of information from which this can be determined:

$v_f = a\cdot t_f\\\implies t_f = \frac{v_f}{a} = \frac{15.2 \frac{m}{s}}{2.36 \frac{m}{s^2}}\approx 6.44 s$

(using vf as final velocity, and tf as final time)

So it takes about 6.44 seconds for the block to move. This allows us to finally calculate the displacement:

$d = \frac{1}{2}at^2=\frac{1}{2}2.36 \frac{m}{s^2}\cdot 6.44^2 s^2 \approx 48.94 m$

and the corresponding work:

$W = F\cdot d=75 N\cdot 48.94 m =3670.5J$

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