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AlekseyPX
3 years ago
8

Person 1 moves an object 10 meters and exerts a force of 50N in 5 seconds. Person 2 moves an object 10 meters and exerts a force

of 50N in 10 seconds. Explain why both people did the same amount of work
Physics
1 answer:
mel-nik [20]3 years ago
7 0

Answer:

Explanation:

Work is a force time the distance moved in the direction of that force, time is not a variable. Provided that the 50 N forces were applied in the same direction, the work done is identical. Assuming both applied force and direction of motion are horizontal W = Fd = 50(10) = 500J.

If the reason that one was slower is because the  second person applied his force at an angle, let's say 60° below the horizontal, then the work done by the second person is 50cos60(10) = 250 J

Time IS a consideration for Power, the RATE of doing work. Provided the force and motion are horizontal, the first person applied twice as much Power as the second person doing an identical amount of work in half the time.

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(a) The equilibrant C for force of vector A and B is 3.43 N.

(b) The equilibrant C for fx of vector A and B is 2.1 N.

(c) The equilibrant  C, for fy of vector A and B is 2.12 N.

<h3>What is equilibrant force?</h3>

An equilibrant force is a single force that will bring other bodies into equilibrium.

<h3>From configuration 1:</h3>

Vector A: mass = 0.2 kg, θ = 20⁰

Vector B: mass = 0.15 kg, θ = 80⁰

Fx = mg cosθ

Fy = mg sinθ

where;

  • m is mass
  • g is acceleration due to gravity

<h3>Vector A</h3>

Force of A due to its weight

F(A) = mg

F(A) = 0.2 x 9.8 = 1.96 N

Fx = (0.2 x 9.8) cos(20) = 1.84 N

Fy = (0.2 x 9.8) sin(20) = 0.67 N

<h3>Resultant force</h3>

R = √(0.67² + 1.84²)

R = 1.96 N

<h3>Vector B</h3>

Force of B due to its weight

F(B) = mg

F(B) = 0.15 x 9.8

F(B) = 1.47 N

Fx = (0.15 x 9.8) cos(80) = 0.26 N

Fy = (0.15 x 9.8) sin(80) = 1.45 N

<h3>Resultant force </h3>

R = √(0.26² + 1.45²)

R= 1.47 N

<h3>Equilibrant  C of vector A and B</h3>

Equilibrant force:

Force, C = 1.96 N + 1.47 N

Force, C = 3.43 N

Equilibrant FX:

Fx, C = Fx(A) + Fx(B)

Fx, C = 1.84 N + 0.26 N = 2.1 N

Equilibrant FY:

Fy, C = Fy(A) + Fy(B)

Fy, C =0.67 N + 1.45 N = 2.12 N

Learn more about equilibrant force here: brainly.com/question/8045102

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