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olga nikolaevna [1]
3 years ago
14

A block of mass m=1.2 kg is held at rest against a spring with a force constant k=730N/m. Initially the spring is compressed a d

istance d. When the block is released, it slides across a surface that is frictionless except for a rough patch of width ????x= 5.0 cm that has coefficient of kinetic friction µk =0.44 Find a general expression of d such that the block’s speed after crossing the rough patch is Vf = 2.3 m/s. Also calculate the value of d. How does the friction effect the d to reach the same Vf?
Physics
1 answer:
igor_vitrenko [27]3 years ago
5 0

Answer:

Explanation:

potential energy of compressed spring

= 1/2 k d²

= 1/2 x 730 d²

= 365 d²

This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .

Kinetic energy after crossing the rough patch

= 1/2 x 1.2 x 2.3²

= 3.174 J

Loss of energy

= 365 d² - 3.174  

This loss is due to negative work done by frictional force

work done by friction = friction force x width of patch

= μmg d ,   μ = coefficient of friction , m is mass of block , d is width of patch

= .44 x 1.2 x 9.8 x .05

= .2587 J

365 d² - 3.174   = .2587

365 d² = 3.4327

d² = 3.4327 / 365

= .0094

d = .097 m

= 9.7 cm

If friction increases , loss of energy increases . so to achieve same kinetic energy , d will have to be increased so that initial energy increases so compensate increased loss .

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PLEASE PLEASE PLEASE A mad scientist places massive amounts of charge on basketball sized aluminum balls. The charge on the ball
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Answer:

4.2 x 10⁷N

Explanation:

Given parameters:

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             q₁  = 3C

              q₂ = 14C

Distance between balls  = 9000m

Unknown:

Force acting on the two balls

Solution:

The force experienced by the two charges is given by coulombs law. It is mathematically expressed as;

                      F  = \frac{k q_{1} q_{2} }{r^{2} }

where k  = 9 x 10⁹Nm²/C²

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Input the variables and solve;

                 

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A stuntman of mass 48 kg is to be launched horizontally out of a spring-
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The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

The Kinetic energy of the stuntman is equal to the elastic potential energy of the spring.

<h3 /><h3>Velocity: </h3>

This is the ratio of displacement to time. The S.I unit of Velocity is m/s.  The velocity of the stuntman can be calculated using the formula below.

⇒ Formula:

  • mv²/2 = ke²/2
  • mv² = ke².................. Equation 1

⇒ Where:

  • m = mass of the stuntman
  • v = velocity of the stuntman
  • k = force constant of the spring
  • e = compression of the spring

⇒ Make v the subject of the equation

  • v = √(ke²/m)................. Equation 2

From the question,

⇒ Given:

  • m = 48 kg
  • k = 75 N/m
  • e = 4 m

⇒ Substitute these values into equation 2

  • v = √[(75×4²)/48]
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Hence, The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

Learn more about velocity here: brainly.com/question/10962624

6 0
2 years ago
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