Question

A box, initially at rest, has 36.7 N of force exerted on it for 2.81 s. If the box has a mass of 7.41 kg, what was its velocity at this time?

Answer 1

Hi there!

We can begin by calculating the acceleration of the box using Newton's Second Law:

`\large\boxed{\Sigma F = ma}`

∑F = Net force (N)

m = mass (kg)

a = acceleration (m/s²)

Rearrange to solve for acceleration:

`(\Sigma F)/(m) = a\\\\(36.7)/(7.41) = 4.95 m/s^2`

Now, we can use the kinematic equation:

`v_f = v_0 + at`

vf = Final velocity (? m/s)

v0 = initial velocity (0 m/s, from rest)

a = acceleration (m/s²)

t = time (s)

Plug in the values:

`v_f = 0 + 4.95(2.81)\\\\v_f = \boxed{13.91 m/s}`

Answer 2

13.91 m/s

Step-by-step explanation:

First we need to find the acceleration:

Acceleration = Force/mass

Acceleration = 36.7N/7.41 kg

Acceleration = 4.95 m/s² (rounded to two decimal places)

Then we find the velocity:

Velocity = Acceleration * Time

Velocity = 4.95 m/s² * 2.81 s

Velocity = 13.91 m/s (rounded to two decimal places)