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Alekssandra [29.7K]
2 years ago
11

A block of mass 5 kg slides down an inclined plane that has an angle of 10. If the inclined plane has no friction and the block

starts at a height of 0.8 m, how much kinetic energy does the block have when it reaches the bottom? Acceleration due to gravity is g = 9.8 m/s2
Physics
1 answer:
Westkost [7]2 years ago
4 0

The kinetic energy of the block  when it reaches the bottom is 39.2 J.

<h3>Kinetic energy of the block at the bottom</h3>

Apply the principle of conservation of energy.

K.E(bottom) = P.E(top)

P.E(top) = mgh

where;

  • m is mass of the block
  • g is acceleration due to gravity
  • h is the vertical height of fall

P.E(top) = 5 x 9.8 x 0.8

P.E(top) = K.E(bottom) = 39.2 J

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

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Answer:

The speed and direction of the two players immediately after the tackle are 3.3 m/s and 53.4° South of West

Explanation:

given information:

mass of fullback, m_{x} = 92 kg

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mass of lineman, m_{y} =110 kg

speed of lineman, v_{y} = 3.6

according to conservation energy,

assume that the collision is perfectly inelastic, thus

initial momentum = final momentum

                            P_{ix} = P_{x}'

                          m₁v₁ = (m₁+m₂)v_{x}'

                             v_{x}' = m₁v₁/(m₁+m₂)

                                  = (92) (5.8)/(92+110)

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                            P_{iy} = P_{y}'

                         m₂v₂ = (m₁+m₂)v_{y}'

                             v_{y}' = m₁v₁/(m₁+m₂)

                                  = (110) (3.6)/(92+110)

                                  = 1.96 m/s

thus,

v' = √v_{x}'²+v_{y}'²

  = 3.3 m/s

then, the direction of the two players is

θ = 90 - tan⁻¹(v_{y}'/v_{x}')

  = 90 - tan⁻¹(1.96/2.64)

  = 53.4° South of West

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