A mass of 10 kg is at a point A on the table. It moved to a point B. If the line joining A and B is horizontal. What is the work
1 answer:
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The center of mass of the two objects is the average position of the parts of the two object system
The center of mass of block <em>A</em>, and block <em>B</em> after displacement of block <em>B</em> is at <u>20 m from block </u><u><em>A</em></u>
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Reason:
The given parameters are;
The position of block A and block B at t = 0 is x = 0
The mass of block A, m₁ = 8 kg
Mass of block B, m₂ = 16 kg
Speed of block <em>A</em> = 0 m/s
Speed of block <em>B</em>, v₂ = 10 m/s
Location of the center of mass of the two object at t = 3 s; Required
Solution;
The location of block <em>A</em>, after 3 s is x₁ = 0 (block A is at rest)
The location of block <em>B</em>, = v₂ × t
The location of block <em>B</em>, after 3 s is x₂ = 10 m/s × 3 s = 30 m
The center of mass of two masses are given as follows;
The center of mass of the two objects is at at the position x = <u>20 m</u> (from block <em>A</em>)
Learn more about the center of mass here:
Answer:
The force of car 3 on car 2 ≈ 1810.82 N
Explanation:
The equation for the change in momentum of the two cars are;
Conservation of linear momentum
150( 2.2 - v) = 265(1.5-v)
150 × 2.2 - 265×1.5 = (150+265)v
150 × 2.2 - 265×1.5 = -67.5 = 415×v
∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East
The impulse of the net force is the amount of momentum change experienced given by the equation;
Impulse force = -
Where;
= The final velocity
= The initial velocity
For the the 265 kg mass, we have;
= 0.1627 m/s
= 1.5 m/s
Which gives the impulse a s F×Δt = 265×0.1627 - 265×1.5 = -354.38 kg·m/s
The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J
Whereby the distance moved in one second is 0.1627 m, we have;
Work done = Force × Distance = Force × 0.1627 = 294.62
Force = 294.62/0.1627 = 1810.82 N.