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a_sh-v [17]
2 years ago
10

Two blocks, of mass m and 2m, are initially at rest on a horizontal frictionless surface. A force F is exerted individually on e

ach block, as shown above. The graph shows how F varies with time t. Which block has the greatest average power provided to it between t = 0 s and t = 3 s?
Physics
1 answer:
Anon25 [30]2 years ago
7 0

The block has the greatest average power provided is bock m.

<h3>What is instantaneous power?</h3>
  • This is the product of force and velocity exerted on an object.

Mathematically instantaneous power is calculated as;

P = Fv

where;

  • F is the applied force
  • v is the velocity

Both blocks (m and 2m) will experience the same force but different velocity.

The smaller block (m) will experience greater velocity.

Thus, the block has the greatest average power provided is bock m.

Learn more about instantaneous power here: brainly.com/question/8893970

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Find velocity vx(t) and coordinate x(t) for a particle of mass m which is subject to the force given by: Fx = F0 e −kt , where F
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Answer:

v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})

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Explanation:

Given that the force of the particle is,

F_{x}=F_{0}e^{-kt}

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m\frac{dv}{dt}= F_{0}e^{-kt}\\\frac{dv}{dt}=\frac{ F_{0}}{m} e^{-kt}\\dv=\frac{ F_{0}}{m}e^{-kt}dt\\ v=\frac{ F_{0}}{-km}e^{-kt}+C

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Now the velocity will become.

v_{x} (t)=\frac{ F_{0}}{-km}e^{-kt}+1+\frac{ F_{0}}{km}\\v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})

And,

\frac{dx}{dt} =1+\frac{ F_{0}}{km}(1-e^{-kt})\\dx=(1+\frac{ F_{0}}{km}(1-e^{-kt}))dt\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+C\\

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0=0+\frac{ F_{0}0}{km}+\frac{ F_{0}t}{k^{2} m}e^{0}+C\\C=-\frac{ F_{0}t}{k^{2} m}

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