Question

A net force of 25 N is applied for 5.7 s to a 12 kg box initially at rest. What is the speed of the box at the end of the 5.7 s interval?

A) 1.8 m/s
B) 12 m/s
C) 3.0 m/s
D) 7.5 m/s
E) 30 m/s

Answer 1

The speed of the box at the end of the 5.7 s interval is approximately 12 m/s after a net force of 25 N is applied, calculated using Newton's second law and the formula for final velocity.

Step-by-step explanation:

To determine the speed of the box at the end of the 5.7 s interval after a net force of 25 N is applied, we use Newton's second law of motion. The acceleration (a) of the box can be found using the formula F = ma, where F is the net force and m is the mass of the object. Once we have the acceleration, we can calculate the final velocity (v) using the formula v = u + at, where u is the initial velocity (which is 0 m/s since the box is initially at rest), a is the acceleration, and t is the time.

Calculating acceleration:
a = F/m = 25 N / 12 kg = 2.08 m/s²

Calculating final velocity:
v = u + at = 0 m/s + (2.08 m/s²)(5.7 s) = 11.856 m/s

Rounding to the nearest whole number, the final velocity of the box is approximately 12 m/s, which corresponds to option B.