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A constant braking force of 10 newtons applied for 5 seconds is used to stop a 2.5-kilogram cart traveling at 20 meters per second. the magnitude of the impulse applied to stop the cart is

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Mathematically it can be expressed as: I = FpΔt. Where F p is the average magnitude of the acting force and Δt = t 2 - t 1, the time lapse in the force acts.
 The magnitude of the impulse applied to stop the cart is
 I = FpΔt = (10N) * (5s) = 50 N.s

The impulse applied to in object produce an equal force to change the leaner momentum of the object.  The magnitude of the impulse applied to stop the cart is 50 N.s.

Impulse:

  • The impulse applied to in object produce an equal force to change the leaner momentum of the object.  
  • SI unit of impulse is the newton second (N⋅s).

The impulse can be calculated as,

[tex]\rm \bold { I = F\times T}[/tex]

Where,

F - force = 10 N

T - time =  5 sec.

put the value in the formula,

[tex]\rm \bold { I = 10 N\times 5 s}\\\\\rm \bold { I = 50 N.s}[/tex]

Hence, we can conclude that the magnitude of the impulse applied to stop the cart is 50 N.s.

To know more about Impulse force, refer to the link:

https://brainly.com/question/24765250

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