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ohaa [14]
3 years ago
6

A(n) 96.1 g ball is dropped from a height of 59.1 cm above a spring of negligible mass.The ball compresses the spring to a maxim

um displacement of 4.75403 cm. The acceleration of gravity is 9.8 m/s^2. Calculate the spring force constant k. Answer in units of N/
Physics
1 answer:
Serhud [2]3 years ago
4 0

Answer:

Explanation:

Mass of ball Is m=96.1g=0.0961kg

Height above spring is 59.1cm

L=0.591m

Extension of the spring is 4.75403cm

e=0.0475403m

Then the distance the ball traveled is H=L+e

H=0.591+0.0475403

H=0.6385403m

Then, the potential energy of the ball is given as

P.E=mgh

P.E=0.0961×9.81×0.6385403

P.E=0.602J

From conservation of energy, energy cannot be created nor destroy but can be transferred from one form to another

Then, the P.E is transferred to the work done by the spring

Then, Work done by spring is given as

W=½ke²

W=P.E=½×k×0.0475403²

0.602=½×k×0.0475403²

k=0.602×2/0.0475403²

k=532.72N/m

The spring constant is 532.72 N/m

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A 5.00-m-long uniform ladder, weighing 200 N, rests against a smooth vertical wall with its base on a horizontal rough floor, a
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Answer:

0.488  m  

Explanation:

If θ be the angle ladder makes with the plane

cos θ = 1.2 / 5

Tan θ = 4.04

Let the height a person of weight 600 N  can climb be h from the ground .

Distance from the base point  where ladder touches the floor  = h / tanθ

= h / 4.04

Total reaction force = total downward force

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Frictional force = μ R

= .2 x 800

= 160 N

Taking moment of force about the point on the ladder  where it  touches the floor  and balancing them

200 x 1.2 x .5 + 600 x   h / tanθ  = μ R x  1.2 / tanθ ( reaction  at the top point of ladder where it touches the wall is  R₁ and

R₁ =μ R   )

= 200 x 1.2 x .5 + 600 x   h / tanθ  = 160 x 1.2 / tanθ

120  - 600 h / 4.04 = 47.52

120 - 47.52 = 600 h / 4.04

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h = 0.488  m  

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