Question

At t= 19.8 seconds an object with mass 2.0 kg was observed to have a velocity of <5, 25, -8> m/s. At t=20.02 seconds its velocity was <16, 24, 24> m/s. What was the average (vector) net force acting on the object?

Answer 1

Explanation:

F = m Δv / Δt

F = (2.0 kg) (<16, 24, 24> m/s − <5, 25, -8> m/s) / (20.02 s − 19.8 s)

F = (2.0 kg) (<11, -1, 32> m/s) / (0.22 s)

F ≈ <100, -9.09, 291> N

Answer 2

The net force acting on the object is (100.1, -9.1, 291.2) N.

The given parameters:

• Mass of the object, m = 2 kg
• Initial time of motion, t₁ = 19.8 s
• Final time of motion, t₂ = 20.02 s
• Initial velocity, u = (5, 25, -8) m/s
• Final velocity, v = (16, 24, 24) m/s

The net force acting on the object is calculated as follows;

$F_{net} = \frac{m\Delta v}{\Delta t} \\\\ F_{net} = \frac{2}{20.02 - 19.8} (16 - 5, \ 24- 15, \ 24 + 8)\\\\ F_{net} = 9.1(11, -1, 32)\\\\ F_{net} = (100.1, \ -9.1, \ 291.2) \ N$

Thus, the net force acting on the object is (100.1, -9.1, 291.2) N.

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