A 6 kg block initially at rest is pulled to East along a horizontal, frictionless surface by a constant horizontal force of 12 N. Find the speed of the block after it has moved 3m.

The Work [tex]W[/tex] done by a Force [tex]F[/tex] refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path. In addition, [tex]W[/tex] is a scalar quantity, and its unit in the International System of Units is the Joule (like energy).

Now, when the applied force is constant and the direction of the force and the direction of motion are parallel, the equation to calculate it is:

[tex]W=F.d[/tex] (1)

Where:

[tex]F= 12 N[/tex] is the force exerted to pull the block to the East

[tex]d= 3 m[/tex] is the distance the block has been moved

[tex]W=(12 N)(3 m)[/tex] (2)

[tex]W=36 Nm= 36 J[/tex] (3)

On the other hand, energy is the ability of matter to produce work in the form of movement, light, heat, among others. In this sense, according to the Work-Kinetic Energy Theorem we have:

[tex]W=\Delta K=K_{f}-K_{o}[/tex] (4)

Where:

[tex]\Delta K[/tex] is the variation in kinetic energy [tex]K[/tex] which is equal to:

[tex]K=\frac{1}{2}mV^{2}[/tex] (5)

Where:

[tex]m=6kg[/tex] is the mass of the block

[tex]V[/tex] is the velocity

Hence, the Work-Kinetic Energy Theorem is written as follows:

[tex]W=\frac{1}{2}mV_{f}^{2} - \frac{1}{2}mV_{o}^{2}[/tex] (6)

Where the initial velocity is zero ([tex]V_{o}=0 m/s[/tex]) since we are told the block is initially at rest:

[tex]W=\frac{1}{2}mV_{f}^{2}[/tex] (7)

So, we have to find the final velocity [tex]V_{f}[/tex], substituting (3) in (7):

[tex]36 J=\frac{1}{2}6 kg V_{f}^{2}[/tex] (8)

[tex]V_{f}=3.46 m/s[/tex] (9) This is the speed of the block after it has moved 3m

b) Acceleration of the block

According to Newton's second law of motion, which relates the force [tex]F[/tex] with the mass [tex]m[/tex] and the acceleration [tex]a[/tex] of an object, we have:

[tex]F=m.a[/tex] (10)

On the other hand, acceleration is the variation of velocity [tex]\Delta V[/tex]in time [tex]\Delta t[/tex]:

[tex]a=\frac{\Delta V}{\Delta t}[/tex] (11)

Then, equation (10) is rewritten as:

[tex]F=m \frac{\Delta V}{\Delta t}[/tex] (12)

Isolating [tex]\Delta t[/tex]:

[tex]\Delta t=\frac{m \Delta V}{F}[/tex] (13)

Since [tex]\Delta V=V_{f}-V_{o}[/tex] and [tex]V_{o}=0[/tex], we have:

[tex]\Delta V=V_{f}=3.46 m/s[/tex]

[tex]\Delta t=\frac{(6 kg) (3.46 m/s)}{12 N}[/tex] (14)

[tex]\Delta t=1.73 s[/tex] (15)

Substituting (15) in (11):

[tex]a=\frac{3.46 m/s}{1.73 s}[/tex] (16)

Finally:

[tex]a=1.99 m/s^{2} \approx 2 m/s^{2}[/tex] This is the acceleration