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A 12.0 kg block is accelerated at 2.70 m/s²

along a horizontal frictionless surface, with the specd increasing from 10.0 m/s to 32.0 m/s : What are (a) the change in the block's mechanical energy and (b) the average rate at which energy is transferred to the biock? What is the instantaneous rate of that transfer when the block's speed is (c) 10.0 m/s and (d) 32.0 m/s ?

User Atif Shafi
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1 Answer

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Final answer:

The change in the block's mechanical energy is 8736 J. The average rate of energy transfer and the instantaneous rates at specific speeds cannot be determined without additional information about time and force.

Step-by-step explanation:

(a) The change in the block's mechanical energy can be found using the equation:


ΔE = 0.5 * m * (v2^2 - v1^2)


where:

  • ΔE is the change in mechanical energy
  • m is the mass of the block (12.0 kg)
  • v1 is the initial velocity (10.0 m/s)
  • v2 is the final velocity (32.0 m/s)


Plugging in the values, we get:


ΔE = 0.5 * 12.0 kg * (32.0 m/s)^2 - 0.5 * 12.0 kg * (10.0 m/s)^2


ΔE = 8736 J


(b) The average rate at which energy is transferred to the block can be found using the equation:


ΔE/Δt


where Δt is the time interval. Since the question does not provide a time interval, we cannot calculate the average rate of transfer.


(c) The instantaneous rate of energy transfer when the block's speed is 10.0 m/s can be calculated using the equation:


Power = Force * velocity


Since no force is mentioned in the question, we cannot calculate the instantaneous rate of energy transfer at this speed.


(d) The instantaneous rate of energy transfer when the block's speed is 32.0 m/s can also be calculated using the equation:


Power = Force * velocity


Again, since no force is mentioned in the question, we cannot calculate the instantaneous rate of energy transfer at this speed.

User Musab Bozkurt
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