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Three dynamics carts have force sensors placed on top of them: i. Cart 1, with mass of 2.2 \ kg,...

Question:

Three dynamics carts have force sensors placed on top of them:

i. Cart 1, with mass of {eq}2.2 \ kg {/eq}, has one sensor,

ii. Cart 2, with mass of {eq}2.5 \ kg {/eq}, has sensors two and then three,

iii. Cart 3, with mass of {eq}1.8 \ kg {/eq}, has sensors four then five.

Each force sensor is tied to a string that connects all three carts together. You use a sixth force sensor to pull the three dynamics carts forward (cart 1 being at the back, cart three at the front, right behind the sixth sensor). The reading on force sensor 2 is {eq}3.3 \ N {/eq}. Assume that the force sensors are light and that there is negligible friction acting on the carts, what is the acceleration of all the carts?

Force:

Let us suppose a puncher car standing on the road and a tractor is pulling it by a rope. In this case, the magnitude of the applied force by the tractor on the car would be equal to the product of the mass and the acceleration gained by the puncher car.

Answer and Explanation:

Given data:

  • Mass of the first car, {eq}m_{1} = 2.2 \ kg {/eq}
  • Mass of the second car, {eq}m_{2} = 2.5 \ kg {/eq}
  • Mass of the third car, {eq}m_{3} = 1.8 \ kg {/eq}
  • Magnitude of force, {eq}F = 3.3 \ \rm N {/eq}

Let the acceleration of the car-system be a.

Thus,

{eq}\begin{align*} a &= \frac{F}{m_{total}}\\ a &= \frac{3.3 }{2.2 + 2.5 + 1.8}\\ a &= 0.51 \ m/s^{2}.\\ \end{align*} {/eq}


Learn more about this topic:

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Force: Definition and Types

from AP Physics 1: Exam Prep

Chapter 5 / Lesson 5
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