Question

A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net work done on the object if its velocity changes to 1 8.00 i ^ 1 4.00 j ^

Answer 1

(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
$v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s$
And so, the kinetic energy of the object is
$K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J$

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
$v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s$
And so the new kinetic energy is
$K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J$

So, the work done on the object is the variation of kinetic energy of the object:
$W=\Delta K=120 J-60 J=60 J$

Answer 2

a. The kinetic energy is equal to 55.5 Joules.

b. The net work done on the object is equal to 64.5 Joules.

Given the following data:

• Initial velocity = $6.00i-2.00j$

• Final velocity = $8.00i+4.00j$

• Mass = 3.00 kg.

For its initial velocity:

$V_i = \sqrt{V_{ix} + V_{iy}} \\\\V_i = \sqrt{6^2 + 1^2}\\ \\V_i = \sqrt{36 + 1}\\\\V_i = \sqrt{37}\;m/s$

For its final velocity:

$V_f = \sqrt{V_{ix} + V_{iy}} \\\\V_f = \sqrt{8^2 + 4^2}\\ \\V_f = \sqrt{64 + 16}\\\\V_f = \sqrt{80}\;m/s$

How to calculate its kinetic energy.

Mathematically, kinetic energy is given by this formula:

$K.E = \frac{1}{2} MV^2$

Where:

• K.E is the kinetic energy.

• M is the mass.

• V is the velocity.

Substituting the given parameters into the formula, we have;

$K.E = \frac{1}{2} \times 3 \times (\sqrt{37}) ^2\\\\K.E = 1.5 \times 37$

K.E = 55.5 Joules.

For the net work done:

$Work\;done =\Delta K.E = \frac{1}{2} M(V^2_f-V^2_i)\\\\Work\;done = \frac{1}{2} \times 3(80-37)\\\\Work\;done = 1.5 \times 43$

Work done = 64.5 Joules.

Read more on kinetic energy here: https://brainly.com/question/1242059