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satela [25.4K]
3 years ago
12

Initially sliding with a speed of 1.9 m/s, a 1.8 kg block collides with a spring and compresses it 0.35 m before coming to rest.

what is the force constant of the spring?
Physics
1 answer:
Alika [10]3 years ago
3 0
Let k =  the force constant of the spring (N/m).

The strain energy (SE) stored in the spring when it is compressed by a distance x=0.35 m is
SE = (1/2)*k*x²
     = 0.5*(k N/m)*(0.35 m)²
     = 0.06125k J

The KE (kinetic energy) of the sliding block is
KE = (1/2)*mass*velocity²
     = 0.5*(1.8 kg)*(1.9 m/s)²
     = 3.249 J

Assume that negligible energy is lost when KE is converted into SE.
Therefore
0.06125k = 3.249
k = 53.04 N/m

Answer:  53 N/m  (nearest integer)

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A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and
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Answer:

a = 1.152s

b = 0.817 m

c = 7.29m/s

Explanation: let the following

From the first equation of linear motion

V = u+at..........1

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t = Time taken

v = Final velocity

a = Acceleration due to gravity = 9.8m/s²

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t = 4/9.8

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From : s = ut+1/2at^2.........2

S = 4×0.408+0.5(-9.8)×0.408^2

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S = 1.632-0.815

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Her highest height above the board is 0.817 m

Total height she would fall is 0.817+1.90 = 2.717 m

From equation 2

s = ut+1/2at^2

2.717 m = 0t+0.5(9.8)t^2

2.717 m = 0+4.9t^2

2.717 m = 4.9t^2

2.717/4.9 = t^2

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t =√0.554

t = 0.744s

Hence, her feet were in the air for 0.744+0.408seconds

= 1.152s

Also recall from equation 1

V= u+at

V = 0+9.8(0.744)

V = 7.29m/s

Hence, the velocity when she hits the water is 7.29m/s

Finally,

a = 1.152s

b = 0.817 m

c = 7.29m/s

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