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The potential energy of a 10 kg mass is given by [tex] -3.98 \times 10^{15} r^{-1} [/tex], where potential energy decreases as [tex] r [/tex] approaches zero.

What is the force on this mass when it is located at [tex] r = 6.38 \times 10^6 \text{ m} [/tex]?

A. 638 N towards [tex] r = 0 [/tex]
B. 638 N away from [tex] r = 0 [/tex]
C. [tex] 62.4 \times 10^8 [/tex] N towards [tex] r = 0 [/tex]
D. 97.8 N away from [tex] r = 0 [/tex]
E. 97.8 N towards [tex] r = 0 [/tex]

Answer :

The force on the mass when it is located at r = 6.38×10⁶ m is approximately 97.8 N away from the reference point.

The potential energy of a mass can be calculated using the equation:

Potential energy = -3.98×10¹âµ / r

where r represents the distance of the mass from its reference point.

In this case, the mass is 10 kg and the potential energy is given as -3.98×10¹âµ / r.

To find the force on the mass when it is located at r = 6.38×10⁶ m, we need to take the derivative of the potential energy equation with respect to r:

Force = d(Potential energy) / d(r)

To calculate the derivative, we treat the -3.98×10¹âµ as a constant:

Force = (-3.98×10¹âµ) * d(1/r) / d(r)

The derivative of 1/r with respect to r is -1/r²:

Force = (-3.98×10¹âµ) * (-1/r²)

Simplifying the expression, we get:

Force = (3.98×10¹âµ) / r²

Now we can substitute the value of r = 6.38×10^6 m into the equation to find the force:

Force = (3.98×10¹âµ) / (6.38×10⁶)²

Calculating this, we get:

Force ≈ 97.8 N away from r=0

Therefore, the force on the mass when it is located at r = 6.38×10⁶ m is approximately 97.8 N away from the reference point.

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