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A 10kg block with an initial velocity of 10 m/s slides 1o m across a horizontal surface and comes to rest. it takes the block 2 seconds to stop. The stopping force acting on the block is about

Respuesta :

Percivle

Answer:

-50N

Explanation:

F=ma=m(Vf-Vi)/t

  • m=10kg
  • Vf=0m/s
  • Vi=10m/s
  • t=2s

F=(10)(-10)/(2)=-50N

So the force acting on the block is -50N, where the negative sign simply tells us that the force is opposite to the direction of movement.

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