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andreev551 [17]
3 years ago
14

A block of 1 kg with a speed 1 m/s hits a spring placed horizontally as shown in the figure. If spring constant is 1000 N/m, fin

d the compression in the spring.
Physics
1 answer:
Evgen [1.6K]3 years ago
8 0

Answer:

0.0316 m

Explanation:

Wok done = Energy change

Work done on the spring = Energy change of the block

  Elastic Potential stored = Kinetic energy of the block

\frac{1}{2} kx^{2} = \frac{1}{2} mv^{2}

                              x = \sqrt{v^{2}\frac{m}{k}  }

                              x = \sqrt{1^{2}\frac{1}{1000}  }

                              x = 0.0316 m

k = spring constant

m = mass of block

v = velocity of the block

x = compression of spring

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Here when an object is placed on the level floor then in that case there are two forces on the object

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so when no force is applied on the box at that time normal force is counter balanced by weight.

Now here it is given that A person tried to lift the box upwards

So now there are two forces on the box

1). Applied force of person

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So now these two forces will counter balance the weight of the crate

So we can write an equation for force balance like

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given that

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Part (a):
For the ball to start falling back to the ground, it has to reach its highest position where its final velocity will be zero.
The equation that we will use here is:
v = u + at where
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u is the initial velocity = 160 m/sec
a is acceleration due to gravity = -9.8 m/sec^2 (the negative sign is because the ball is moving upwards, thus, its moving against gravity)
t is the time that we want to find.
Substitute in the equation to get the time as follows:
v = u + at
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t = 160/9.8 = 16.3265 sec
Therefore, the ball would take 16.3265 seconds before it starts falling back to the ground

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First, we will get the total distance traveled by the ball as follows:
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s = 0.5(160+0)*16.3265
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The equation that we will use to solve this part is:
v^2 = u^2 + 2as where
v is the final velocity we want to calculate
u is the initial velocity of falling = 0 m/sec (ball starting falling when it reached the highest position, So, the final velocity in part a became the initial velocity here)
a is acceleration due to gravity = 9.8 m/sec^2 (positive as ball is moving downwards)
s is the distance covered = 1306.12 meters
Substitute in the above equation to get the final velocity as follows:
v^2 = u^2 + 2as
v^2 = (0)^2 + 2(9.8)(1306.12)
v^2 = 25599.952 m^2/sec^2
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