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3 years ago

How does changing the lengthy but not the height of an inclined plane affect the work done to lift a load? PLZ HELP ME NOW'

Physics

1 answer:

Serhud [2]3 years ago

4 0

Answer: Work Done would remain same.

Let us assume that the velocity is constant while taking the load up the inclined plane. Then, the kinetic energy would remain the same. This is because kinetic energy is dependent on velocity $(K.E.=\frac{1}{2}mv^2)$ . If that is constant, the kinetic energy would remain same. The potential energy is dependent on the height $(P.E.=mgh)$ . If the height is changed, then potential energy varies. In the question, it is mentioned that without changing the height, the length of the inclined plane is changed. Therefore, the potential energy would be same as before.

We know, work done is equal to potential energy plus kinetic energy. Since there is no change in any of these, the required work done would not change.

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A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 20.1 N; when it is completely i

Paha777 [63]

Answer:

A) $V = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}$

B) $d = 4181.49 kg/m^{3} = 4.18 g/cm^{3}$

Explanation:

A) Using the Archimedes' force we can find the weight of water displaced:

$W_{d} = W_{a} - W_{w}$

Where:

$W_{a}$ : is the weight of the block in the air = 20.1 N

$W_{w}$ : is the weight of the block in the water = 15.3 N

$W_{d} = W_{a} - W_{w} = 20.1 N - 15.3 N = 4.8 N$

Now, the mass of the water displaced is:

$m = \frac{W_{d}}{g} = \frac{4.8 N}{9.81 m/s^{2}} = 0.49 kg$

The volume of the block can be found using the mass of water displaced and the density of the water:

$V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}$

B) The density of the block can be found as follows:

$d = \frac{W_{a}}{g*V} = \frac{20.1 N}{9.81 m/s^{2}*4.92 \cdot 10^{-4} m^{3}} = 4181.49 kg/m^{3} = 4.18 g/cm^{3}$

I hope it helps you!

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