Consider two point charges located on the x axis: one charge, q1 = -11.5nC , is located at x1 = -1.685m ;the second charge, q2 =
40.0nC , is at the origin (x=0.0000). What is the net force exerted by these two charges on a third charge q3 = 55.0nC placed between q1 and q2 at x3 = -1.195m ?
Your answer may be positive or negative, depending on the direction of the force.
The formula to compute the electrostatic force between two point charges was introduced by Charles Coulomb and it's expressed as
Where q1 and q2 are the magnitudes of the charges, r is the distance between them and k is a proportionality constant
The data provided in the problem is
The distance between the charge 1 and 3 is
The distance between charges 2 and 3 is
Now, let's compute the force exerted by q1 on q3
This value is the scalar magnitude of the force, we'll take care of the directions later. The force by q2 on q3 is
Given that q1 is negative and q3 is positive, q1 attracts q3 to the left, so F13 is in the negative direction. The charge q2 is positive and repels q3 to the left, thus F23 is also negative. This leads us to compute the total force on q3 as
Given there are three blocks of masses , and (ref image in attachment)
When all three masses move together at an acceleration a, the force F is given by
F = ( + + ) *a ................(equation 1)
Also it is given that does not move with respect to , which gives tension T is exerted on pulley by only, Hence tension T is
T = *a ..........(equation 2)
There is also also tension exerted by . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by
T = ................(equation 3)
From equation 2 and 3, we get
*a =
Squaring both sides we get
* = * (+)
* = ( * )+ ( *)
( - ) * = *
= */( - )
Taking square root on both sides, we get acceleration a
a = *g/()
Hence substituting the value of a in equation 1, we get
The equation (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.
The horizontal momentum is given by:
Where:
m₁: is the mass of the lab cart = 15 kg
m₂: is the <em>mass </em>of the object dropped = 2 kg
: is the initial velocity of the<em> lab cart </em>
: is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)
: is the final velocity of the<em> lab cart </em>
: is the <em>final velocity</em> of the <em>object </em>
Then, the horizontal momentum is:
When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:
Therefore, the equation represents the horizontal momentum (option 3).