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Svetllana [295]
3 years ago
7

What is work done by a person in moving a block of mass 2kg on a frictionless floor up to a distance of 5m?

Physics
1 answer:
Sav [38]3 years ago
8 0
If the floor is truly frictionless, then ANY force, no matter how small, applied for
ANY time, no matter how short, will set the block in motion. The force can then
be taken away, because on a frictionless surface, NO force is required in order
to keep the block moving.

The work done is equal to the kinetic energy you want the block to have. That
can be as much or as little as you feel like giving it.
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7 0
4 years ago
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Friction in machines is commonly reduced by using what?
Dominik [7]
I am not sure if this is what your looking for but most machines use oil 
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4 years ago
Which is the correct interpretation of the formula mgf2? the lattice contains 1 magnesium atom for every 2 fluorine atoms. the l
mixer [17]

Answer: the lattice contains 1 magnesium ion for every 2 fluoride ions.

MgF_2  is an ionic compound. It contains positive ions of Magnesium and  and negative ions of Fluorine. To balance the +2 positive charge of one magnesium ion, two (-1) negatively charged Fluorine ions are present. Hence, The correct interpretation of MgF_2  is that the lattice contains 1 Magnesium ion for every 2 Fluoride ions.

7 0
3 years ago
Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three.
SpyIntel [72]

Answer:

The speed decreases 75%.

Explanation:

  • Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
  • For the first collission, only mass 1 is moving before it, so we can write the following equation:

       p_{i} = p_{f} = m*v_{o}    (1)

  • Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:

       p_{f1} = 2*m*v_{1}    (2)

  • From (1) and (2) we get:
  • v₁ = v₀/2  (3)
  • Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:

       p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2}  = m*v (4)

  • Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:

        p_{2} = 3*m*v_{2}  (5)

  • From (4) and (5) we get:
  • v₂ = v₀/3  (6)
  • Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:

      p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3}  = m*v (7)

  • Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:

       p_{3} = 4*m*v_{3}  (8)

  • From (7) and (8) we get:
  • v₃ = v₀/4
  • This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.
5 0
3 years ago
You and a friend each carry identical boxes from the first floor of a building to a room located on the second floor, farther do
melamori03 [73]

Answer:

Work done in both the cases will be same

Explanation:

As we know that the work done against gravity is given as

W = F_g .d

here we know that gravitational force is a conservative force and the work done against gravitational force is independent of the path

So here the work done by person to move the object between two different heights will be independent of the path they choose

So for the first person and second person will be same in both the cases because the height through which the boxes are transferred will be same in both the cases

8 0
3 years ago
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