Problem: A 5.50-kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t) = (2.80 m/s)t + (0.61 m/s3)t3.What is the magnitude of the force F when t = 4.10 s?

FREE Expert Solution

In this problem, we are required to determinalso don't think it's a great idea toe the magnitude of the force on an object given the position as a time-varying function

Since we have the position as a function of time, we know this is a calculus problem. 

A diagram like this one can help you remember the relationships between the variables:

PVddtdtA,FF=ma

Since we're looking for the magnitude of the force exerted on the object, we'll have to find an expression for acceleration, a(t). The a(t) expression is obtained by differentiating the position function, y(t), twice.  Then, we'll multiply the a(t) function by the mass, m, to get the force expression.

The steps needed to solve this problem are simple and straight forward:

  1. Differentiate the position function, y(t), twice to get the acceleration function a(t). Remember, differentiating the position function, y(t), once gives the velocity function, v(t).
  2. Use the equation ma to find a function for the force
  3. Calculate the force at a specific time of interest.

In Step 1, we'll need to remember the power rule of differentiation:

ddt(tn)=ntn-1 

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Problem Details

A 5.50-kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t) = (2.80 m/s)t + (0.61 m/s3)t3.
What is the magnitude of the force F when t = 4.10 s?

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