To calculate the final speed of the body, we can use the formula:
\(v_f = v_i + at\)
Where:
\(v_f\) = final velocity
\(v_i\) = initial velocity
\(a\) = acceleration
\(t\) = time
Given:
\(m = 2.0 kg\)
\(v_i = 40. m/s\)
\(F = 10. N\)
\(t = 5.0 s\)
We can calculate the acceleration (\(a\)) first using Newton's Second Law:
\(F = ma\)
\(a = F/m\)
\(a = 10. N / 2.0 kg\)
\(a = 5.0 m/s^2\)
Now, we can substitute the values into the initial formula:
\(v_f = 40. m/s + (5.0 m/s^2)(5.0 s)\)
\(v_f = 40. m/s + 25.0 m/s\)
\(v_f = 65.0 m/s\)
Therefore, the final speed of the body is 65 m/s. So, the correct answer is:
65 m/s
A 2.0-kilogram body is initially traveling at a velocity of 40. meters per second east. If a constant force of 10. newtons due east is applied to the body for 5.0 seconds, what is the final speed of the body?
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1 point
15 m/s
25 m/s
65 m/s
130 m/s
1 answer