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).
