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GuDViN [60]
4 years ago
7

In a system with only a single force acting upon a body, what is the relationship between the change in kinetic energy and the w

ork done by the force?
Physics
1 answer:
Makovka662 [10]4 years ago
6 0

Answer: W.D = 1/2mv^2

Explanation:

If an external force or a single force is acting on a body. Just like the first law of thermodynamics, the force acting on the body will cause work done on the system.

Work done = force × distance

And the work done on the body will cause the molecules of the body to experience motion and thereby producing kinetic energy.

The work done will be converted to kinetic energy.

W.D = 1/2mv^2

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The best method is using the jacket

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3 years ago
How many miles will a car drive on 25 L of gasoline if the car average mileage is 60 km/gal. provide answer with correct signifi
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Distance traveled by car in 25 L IS 396.15 km

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4 0
3 years ago
on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker
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<u>Momentum</u> 
- a vector quantity; has both magnitude and direction
- has the same direction as object's velocity
- can be represented by components x & y.

Find linebacker momentum given m₁ = 120kg, v₁ = 8.6 m/s north
P₁ = m₁v₁
P₁ = (120)(8.6)
[ P₁ = 1032 kg·m/s ] = y-component, linebacker momentum

Find halfback momentum given m₂ = 75kg, v₂ = 7.4 m/s east
P₂ = m₂v₂
P₂ = (75)(7.4)
[ P₂ = 555 kg·m/s ] = x-component, halfback momentum

Find total momentum using x and y components.
P = √(P₁)² + (P₂)²
P = √(1032)² + (555)²
[[ P = 1171.77 kg·m/s ]] = magnitude 

!! Finally, to find the magnitude of velocity, take the divide magnitude of momentum by the total mass of the players.
P = mv
P = (m₁ + m₂)v
1171.77 = (120 + 75)v      <em>[solve for v]</em>
<em />v = 1171.77/195
v = 6.0091 ≈ 6.0 m/s

If asked to find direction, take inverse tan of x and y components.
tanθ = (y/x)
θ = tan⁻¹(1032/555)
[ θ = 61.73° north of east. ]

The magnitude of the velocity at which the two players move together immediately after the collision is approximately 6.0 m/s.
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