A 9 kg object is given a displacement
∆~s = (4 m) ý + (3 m)ˆ J + (−4 m) k
along a straight line. During the displace-
ment, a constant force of
~ F = (1 N) ý + (−5 N) J + (7 N) k
acts on the object.
Find the work done by ~ F for this displace-
ment.
Answer in units of J.
Find the component of ~ F in the direction of
this displacement.
Answer in units of N.
To find the work done by the force ~F for the given displacement, we can use the formula:
Work = Force * Displacement * cos(theta)
where theta is the angle between the force and the displacement vectors.
First, we need to find the angle between the force and the displacement.
The dot product of two vectors A and B is given by:
A · B = |A| * |B| * cos(theta)
where |A| and |B| are the magnitudes of vectors A and B respectively.
Let's calculate the dot product of ~F and the displacement vector ∆~s:
~F · ∆~s = (1 N) * (4 m) + (-5 N) * (3 m) + (7 N) * (-4 m)
= 4 Nm - 15 Nm - 28 Nm
= -39 Nm
Now, let's calculate the magnitudes of ~F and ∆~s:
|~F| = sqrt((1 N)^2 + (-5 N)^2 + (7 N)^2) = sqrt(1 N + 25 N + 49 N) = sqrt(75 N) = 5√3 N
|∆~s| = sqrt((4 m)^2 + (3 m)^2 + (-4 m)^2) = sqrt(16 m + 9 m + 16 m) = sqrt(41 m)
Now we can calculate the angle theta:
cos(theta) = (~F · ∆~s) / (|~F| * |∆~s|)
cos(theta) = (-39 Nm) / [(5√3 N) * sqrt(41 m)]
cos(theta) = -39 / [5√3 * sqrt(41)]
Theta = acos(cos(theta)) = acos(-39 / [5√3 * sqrt(41)])
Now we can calculate the work done by F:
Work = ~F * ∆~s * cos(theta)
Work = (1 N, -5 N, 7 N) * (4 m, 3 m, -4 m) * cos(theta)
Work = 1 N * 4 m * cos(theta) + -5 N *3 m * cos(theta) + 7 N * -4 m *cos(theta)
Work = 4cos(theta) -15cos(theta) -28cos(theta)
Work = -39*cos(theta) J
The answer for the work done by F for this displacement is -39*cos(theta) J.
To find the component of ~F in the direction of this displacement, we can calculate the dot product of ~F and the unit vector in the direction of the displacement. The unit vector can be found by dividing the displacement vector by its magnitude.
Let's calculate the unit vector in the direction of the displacement:
unit vector = (4 m, 3 m, -4 m) / sqrt((4 m)^2 + (3 m)^2 + (-4 m)^2)
unit vector = (4 m, 3 m, -4 m) / sqrt(16 m + 9 m + 16 m) = (4/√41, 3/√41, -4/√41)
Now, let's calculate the dot product of ~F and the unit vector:
~F · unit vector = (1 N, -5 N, 7 N) · (4/√41, 3/√41, -4/√41)
~F · unit vector = (1 N) * (4/√41) + (-5 N) * (3/√41) + (7 N) * (-4/√41)
~F · unit vector = (4/√41) - (15/√41) - (28/√41)
~F · unit vector = (-39/√41) N
Therefore, the component of ~F in the direction of this displacement is (-39/√41) N.