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il63 [147K]
3 years ago
15

A spring with a force constant of 5.3 n/m has a relaxed length of 2.60 m. when a mass is attached to the end of the spring and a

llowed to come to rest, the vertical length of the shaik (rs53237) – work, energy, power – theisen – (20167) 2 spring is 3.62 m. calculate the elastic potential energy stored in the spring. answer in units of j
Physics
1 answer:
olganol [36]3 years ago
6 0
This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant. 
<span> U = (1/2)kx^2 
</span><span> U = (1/2)(5.3)(3.62-2.60)^2 
</span> U = <span> <span>2.75706 </span></span>J
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Lelechka [254]

Answer:1.902 m

Explanation:

Given

height of apartment=1.5 m

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So

s=u_1t+\frac{gt^2}{2}

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u_1=6.11 m/s

i.e. if ball is dropped from top its velocity at window is 6.11 m/s

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v^2-u^2=2as

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here u=0

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3 years ago
A particle covers 10m distance in 2sec. if final velocity is 8m/s, find initial velocity​
klasskru [66]

Explanation:

It is given that a particle covers 10m in first 5s and 10m in next 3s. so using the equation of motion

Case I

s=ut+

2

1

at

2

10=5u+

2

1

a(5)

2

20=10u+25a

4=2u+5a..............(1)

Case 2

In next 3s the particle covers more 10m distance. So

20=8u+

2

1

a(8)

2

5=2u+8a.........(2)

On solving equation (1) and (2)

4=2u+5a

5=2u+8a

a=

3

1

m/s

2

Put the value of a in equation (1)

u=

6

7

m/s

Now to find distance in next 10 s. total time will be 10s

s=

6

7

×10+

2

1

×

3

1

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2

s=28.33m

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s=28.33−20=8.33m

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