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A 5 kg box is lying at rest on a frictionless surface. If you push it with a constant net force of 10 N, how far will it travel over the first 4 s

Respuesta :

asuwafohjames

The distance traveled by the box over the first 4 s  is 16 m.

To solve the problem above,  First, we apply newton's Fundamental equation of force.

Newton's fundamental equation of force:

F = ma...................... Equation 1

Where

  • F = Force
  • m = mass of the box
  • a = acceleration

make "a " the subject of the equation

a = F/m...................... Equation 2

From the question,

Given:

  • F = 10 N
  • m = 5 kg

Substitute these values into equation 2

a = 10/5

a = 2 m/s²

Finally, we use the equation of motion to calculate the distance it will travel over the given period of time.

Equation of motion

s = ut+at²/2........................... Equation 3

Where:

  • s = total distance traveled
  • t = time
  • u = initial velocity

Given:

  • t = 4 s
  • u = 0 m/s (at rest)
  • a = 2 m/s²

Substitute these values into equation 3

  • s = 0(4)+(2×4²)/2
  • s = 0+32/2
  • s = 16 m

Hence, the distance traveled by the box over the first 4 s  is 16 m

Learn more about Newton's fundamental equation here: https://brainly.com/question/13370981

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