Calculate the work done by gravity as a 10 kg object is moved from point A = (0,0,0) to point B = (1,2,0). We are given s = (x sub f - x sub I)x(hat) + (y sub f - y sub I)y(hat) + (z sub f - z sub I)z(hat). Sorry if that doesn't make sense. I think I have to plug in the numbers into the equation but I don't know how to calculate the work done with the 10 kg object.
All that matters here is how far the object is lifted. (increase in U = m g h)
If z is the up direction, the work done is zero. If y is the up direction, U = 10*9.81*2
By the way usually the vector is written:
S = (X2-X1)i + (Y2-Y1)j + (Z2-Z1)k
Where I, j and k are unit vectors in the x, y and z directions.
If Z is up, the work done is zero going from z = 0 to z = 0
since the force vector is in the -k direction
dU = F dot dS
= [ 0 i + 0 j - m g k] dot [ dX i + dY j + dZ k]
= m g dZ (which is a scalar change in potential)
To calculate the work done by gravity as the 10 kg object is moved from point A to point B, we can use the formula:
Work = Force x Displacement x cos(theta)
In this case, the force is the force due to gravity acting on the object, which can be calculated using the formula:
Force = mass x gravity
where mass is 10 kg (given) and gravity is approximately 9.8 m/s^2 (standard value).
The displacement vector, s, between point A and point B is given by:
s = (x_f - x_i) x-hat + (y_f - y_i) y-hat + (z_f - z_i) z-hat
Here, (x_f, y_f, z_f) represent the coordinates of point B and (x_i, y_i, z_i) represent the coordinates of point A.
Given that point A is (0, 0, 0) and point B is (1, 2, 0), we can substitute these values into the displacement equation:
s = (1 - 0) x-hat + (2 - 0) y-hat + (0 - 0) z-hat
= 1 x-hat + 2 y-hat + 0 z-hat
= x-hat + 2 y-hat
Now, we have the values for force and displacement. To find the angle theta between the force vector and the displacement vector, we can use the following trigonometric relationship:
cos(theta) = (Force dot s) / (|Force| |s|)
where "." denotes the dot product and "|" represents the magnitude.
In our case, the dot product of force and displacement is simply:
Force dot s = |Force| |s| cos(theta)
since the force and displacement vectors are collinear. Therefore, we can simplify the equation to:
cos(theta) = 1
as the magnitude of force and displacement vectors will cancel out.
As cos(theta) = 1, we can write:
Work = Force x Displacement x cos(theta)
= (mass x gravity) x |s| x 1
= mass x gravity x |s|
Substituting the known values:
Work = 10 kg x (9.8 m/s^2) x |s|
To calculate the magnitude of s, we can use the distance formula:
|s| = sqrt((x_f - x_i)^2 + (y_f - y_i)^2 + (z_f - z_i)^2)
In our case:
|s| = sqrt((1 - 0)^2 + (2 - 0)^2 + (0 - 0)^2)
= sqrt(1 + 4 + 0)
= sqrt(5)
Therefore, the work done by gravity as the object is moved from point A to point B is:
Work = 10 kg x (9.8 m/s^2) x sqrt(5)