Questions: A 78.0 kg object is traveling due north at 15.2 m / s when it begins to experience a constant net force of 1.72 kN toward the south. What will its velocity be 0.75 s later? Multiple Choice 1.30 m / s, south. 16.5 m / s, south 1.30 m / s, north. 18.8 m / s, south 18.8 m / s, north. 16.5 m / s, north

A 78.0 kg object is traveling due north at 15.2 m / s when it begins to experience a constant net force of 1.72 kN toward the south. What will its velocity be 0.75 s later?

Multiple Choice
1.30 m / s, south.
16.5 m / s, south
1.30 m / s, north.
18.8 m / s, south
18.8 m / s, north.
16.5 m / s, north

Transcript text: A 78.0 kg object is traveling due north at $15.2 \mathrm{~m} / \mathrm{s}$ when it begins to experience a constant net force of 1.72 kN toward the south. What will its velocity be 0.75 s later? Multiple Choice $1.30 \mathrm{~m} / \mathrm{s}$, south. $16.5 \mathrm{~m} / \mathrm{s}$, south $1.30 \mathrm{~m} / \mathrm{s}$, north. $18.8 \mathrm{~m} / \mathrm{s}$, south $18.8 \mathrm{~m} / \mathrm{s}$, north. $16.5 \mathrm{~m} / \mathrm{s}$, north

Solution

Solution Steps

Step 1: Determine the acceleration

First, we need to find the acceleration of the object. The net force $ F $ is given as 1.72 kN (which is 1720 N), and the mass $ m $ is 78.0 kg. Using Newton's second law, $ F = ma $, we can solve for the acceleration $ a $:

\[ a = \frac{F}{m} = \frac{1720 \, \text{N}}{78.0 \, \text{kg}} = 22.0513 \, \text{m/s}^2 \]

Step 2: Calculate the change in velocity

Next, we calculate the change in velocity $ \Delta v $ over the time interval $ t = 0.75 \, \text{s} $. The object is experiencing a constant acceleration in the direction opposite to its initial velocity (southward):

\[ \Delta v = a \cdot t = 22.0513 \, \text{m/s}^2 \cdot 0.75 \, \text{s} = 16.5385 \, \text{m/s} \]

Step 3: Determine the final velocity

The initial velocity $ v_0 $ is 15.2 m/s northward. Since the acceleration is southward, it will reduce the northward velocity. The final velocity $ v_f $ is given by:

\[ v_f = v_0 - \Delta v = 15.2 \, \text{m/s} - 16.5385 \, \text{m/s} = -1.3385 \, \text{m/s} \]

The negative sign indicates that the final velocity is southward.