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Natalka [10]
3 years ago
15

A massless spring with force constant ????=200N/m hangs from the ceiling. A 2.0-kg block is attached to the free end of the spri

ng and released. If the block falls 17 cm before starting back upwards, how much work is done by friction during its descent?
Physics
1 answer:
Makovka662 [10]3 years ago
8 0

Answer:

-0.4454 Joules

Explanation:

m = Mass of block = 2 kg

h = Height of extension = 17 cm = x

g = Acceleration due to gravity = 9.81 m/s²

Potential energy of the spring

P=mgh\\\Rightarrow P=2\times 9.81\times 0.17\\\Rightarrow P=3.3354\ J

The kinetic energy of the spring

K=\frac{1}{2}mx^2\\\Rightarrow K=\frac{1}{2}\times 200\times 0.17^2\\\Rightarrow K=2.89\ J

In this system as the potential and kinetic energy is conserved from work energy equivalence we get

W=P-K\\\Rightarrow W=2.89-3.3354\\\Rightarrow W=-0.4454\ J

The work done by friction is -0.4454 Joules

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What net force would be necessary to cause a block of wood with a mass of 2.5 kg to accelerate at a rate of 3.0 m/s2
charle [14.2K]

Answer:

<h2>7.5 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

mass = 2.5 kg

acceleration = 3.0 m/s²

We have

force = 2.5 × 3.0 = 7.5

We have the final answer as

<h3>7.5 N</h3>

Hope this helps you

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3 years ago
How would you describe the behavior of particles in a solid?
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How would you describe the behavior of particles in a solid?
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The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm
Alika [10]

Answer:

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Explanation:

We have given the force,  F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by F=kx

So 63.5=k\times 0.0531

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J

8 0
3 years ago
Two tuning forks of frequency 480 hz and 484 hz are struck simultaneously. what is the beat frequency resulting from the two sou
umka21 [38]
Ans: Beat frequency = f_b = 4Hz

Explanation: 
The beat frequency is equal to the absolute value of the difference in frequency of the two waves. In other words, the number of beats per second is equal to the difference in frequency. It is due to the destructive and constructive interference. <span>According to this interference, sound will be soft or loud.

Hence. the formula is:
</span>Beat frequency = f_b = |f_2 - f_1|
<span>
Since,
</span>f_1 = 480Hz
f_2 = 484Hz

Therefore,
Beat frequency = f_b = |484 - 480|

=> Beat frequency = f_b = 4Hz
-i
7 0
3 years ago
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