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A boy with a mass of 25 kg is sitting in a red wagon of mass 8.5 kg which is at rest. His friend begins pulling him forward, accelerating him with a constant force for 2.35 s to a speed of 1.8 m/s. Calculate the force with which the wagon was being pulled.

4.2 N

25.7 N

19.1 N

6.5 N

Sagot :

MDKPfromIDNidn

So, the force that given when the wagon was being pulled is approximately 19.1 N (C).

Introduction

Hi ! For intermesso, this question will adopt a lot about the relationship of impulse to change in momentum. Impulse is the total force applied in a certain time interval. Impulses can cause a change of momentum, because momentum itself is a mass that is affected by the velocity of an object. We know that velocity is a vector quantity easy to change its direction. The relationship between impulse and change in momentum is formulated by :

[tex] \sf{I = \Delta p} [/tex]

[tex] \sf{F \cdot \Delta t = (m \cdot v') - (m \cdot v)} [/tex]

[tex] \boxed{\sf{\bold{F \cdot \Delta t = m (v' -v)}}} [/tex]

With the following condition :

  • I = impulse that given (N.s)
  • [tex] \sf{\Delta p} [/tex] = change of momentum (kg.m/s)
  • F = force that given (N)
  • m = mass of the object (kg)
  • v = initial velocity (m/s)
  • v' = final velocity (m/s)
  • [tex] \sf{\Delta t} [/tex] = interval of the time (s)

Problem Solving

We know that :

  • m = mass of the object = 25 kg
  • v = initial velocity = 0 m/s
  • v' = final velocity = 1.8 m/s
  • [tex] \sf{\Delta t} [/tex] = interval of the time = 2.35 s

What was asked :

  • F = force that given = ... N

Step by step :

[tex] \sf{F \cdot \Delta t = m (v' -v)} [/tex]

[tex] \sf{F \cdot 2.35 = 25 (1.8 - 0)} [/tex]

[tex] \sf{F = \frac{25 (1.8)}{2.35}} [/tex]

[tex] \boxed{\sf{F = 19.15 \: N \approx 19.1 \: N}} [/tex]

Conclusion

So, the force that given when the wagon was being pulled is approximately 19.1 N (C).

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