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A 4.0-kg block is pushed up a 36 incline by a force of magnitude P applied parallel to the incline. When P is 31 N, it is observed that the block moves up the incline with a constant speed. What value of P would be required to lower the block down the incline at a constant speed?

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usmanbiu

Answer:

15.1 N

Explanation:

mass of block (m) = 4 kg

angle of inclination = 36 degrees

applied force (P) = 31 N

acceleration due to gravity (g) = 9.8 m/s^{2}

since the block is moving at a constant speed, it means the acceleration is 0 and therefore the summation of all the forces acting on the body is 0

therefore

P - f - mgsinθ = 0

where

  • P = applied force
  • f = frictional force
  • m = mass
  • g = acceleration due to gravity
  • θ = angle of inclination

when P = 31 N and the block is pushed upward

31 - f - (4 x 9.8 x sin 36) = 0

f = 7.96 N

now that we have the value of the frictional force we can find P required to lower the block, our equation becomes p + f - mgsinθ = 0 since the block is to be lowered

P + f - mgsinθ = 0

P = mgsinθ - f

P = (4 x 9.8 x sin 36) - 7.96 = 15.1 N

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