A 3.0 kg object is driven along an x axis by a variable force that is directed along that axis. Its position is given by `x=4.0" m"+(5.0" m"//"s")t+kt^(2)-(3.0" m"//"s"^(3))t^(3)`, where x is measured in meters and t in seconds. The factor k is a constant. At t = 4.0 s, the force on the particle has a magnitude of 37 N and is in the negative direction of the axis. Find the value of k.

Explanation

To find the value of the constant \( k \) in the given position function of the object, we will follow these steps:

### Step 1: Write down the position function
The position of the object is given by:
\[ x(t) = 4.0 \, \text{m} + (5.0 \, \text{m/s}) t + kt^2 - (3.0 \, \text{m/s}^3) t^3 \]

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