A 3.00 kg object has a velocity (5.20 ihat - 2.40 jhat) m/s. (Note: From the definition of the dot product, v2 = vvec·vvec.)
(a) What is its kinetic energy at this time?
Could some one just please show me how to convert the velocity vector into the "regular" form?
Thank You
The vector form you were given is fairly standard. When you take the dot product of the vectopr with itself, you get the scalar magnitude squared, which is what you want for the kinetic energy,
|V|^2 = (5.2)^2 + (2.4)^2
Note that is the same thng as applying the Pythagorean Therem to the two components.
Certainly! To convert a velocity vector into "regular" form, we need to extract the magnitude (speed) and direction of the velocity.
Given the velocity vector v = (5.20 ihat - 2.40 jhat) m/s, we can see that the ihat component represents the motion in the x-direction, and the jhat component represents the motion in the y-direction.
The magnitude (speed) of the velocity can be calculated using the Pythagorean theorem. It is given by:
|v| = √(v_x^2 + v_y^2)
Plugging in the values from the velocity vector, we have:
|v| = √(5.20^2 + (-2.40)^2)
= √(27.04 + 5.76)
= √32.80
≈ 5.729 m/s (rounded to three decimal places)
To determine the direction of the velocity, we can use trigonometry. The direction is usually measured as an angle with respect to the positive x-axis, also known as the angle of inclination (θ).
The angle θ can be calculated using the inverse tangent (arctan) function:
θ = arctan(v_y / v_x)
Plugging in the values from the velocity vector, we have:
θ = arctan((-2.40) / 5.20)
≈ -24.87° (rounded to two decimal places)
Note that the negative sign indicates that the velocity vector is pointing in the fourth quadrant.
Now that we have converted the velocity vector into its magnitude and direction, we can proceed to calculate the object's kinetic energy.
The kinetic energy (K.E.) of an object is given by:
K.E. = (1/2) * m * v^2
where m is the mass of the object and v is the magnitude of the velocity.
In this case, the mass of the object is given as 3.00 kg, and the magnitude of the velocity is approximately 5.729 m/s.
Plugging in the values, we have:
K.E. = (1/2) * 3.00 kg * (5.729 m/s)^2
≈ 49.334 J (rounded to three decimal places)
Therefore, the kinetic energy of the object is approximately 49.334 Joules.