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Zepler [3.9K]
2 years ago
5

An object has k.e. of 10J at a certain instant. If it is acted on by an opposing force of 5N, which of the number A to E below i

s the farthest distance it travels in metres before coming to rest?
A - 2
B - 5
C - 10
D - 20
E - 50
Physics
1 answer:
Solnce55 [7]2 years ago
8 0

Answer:

d = 2 [m].

Explanation:

To solve this problem we must use the principle of conservation of energy, where the mechanical energy in a state plus the work done on the body, must be equal to the mechanical energy in the state E. This can be easily represented in the following equation.

E_{A}+W_{A-E}=E_{E}

where:

Ea = Mechanical energy in A [J]

Wa-e = Work among states a and e [J]

Ee = Mechanical energy in E [J].

The key to being able to understand this problem is that in state A, we only have kinetic energy, while the energy in state E is equal to zero (there is no movement). The work is equal to the product of force by distance, as work acts in the opposite direction to movement, this has a negative sign.

10 - F*d = 0\\5*d = 10\\d = 10/5\\d = 2 [m]

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We know that the Delta E + W(Work done by non-conservative forces) = 0 (change of energy)

In here, the non-conservative force is the friction force where f = uN (u =kinetic friction coefficient) 

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OlgaM077 [116]
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olchik [2.2K]

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Explanation:

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ohaa [14]

Answer:

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