At the instant the displacement of a 2.00 kg object relative to the origin is d = (2.00 m) ihat + (4.00 m) jhat - (3.00 m) khat, its velocity is v = - (6.20 m/s) ihat + (2.70 m/s) jhat + (2.60 m/s) khat, and it is subject to a force F = (6.00 N) ihat - (7.70 N) jhat + (4.10 N) khat.
(a) Find the acceleration of the object.
The jhat, khat, and the ihat.
(b) Find the angular momentum of the object about the origin.
The jhat, khat, and the ihat.
(c) Find the torque about the origin acting on the object.
The jhat, khat, and the ihat.
(d) Find the angle between the velocity of the object and the force acting on the object.
The jhat, khat, and the ihat.
(a) To find the acceleration of the object, we can use Newton's second law of motion: F = m * a, where F is the net force acting on the object, m is the mass of the object, and a is the acceleration.
Given:
Mass of the object, m = 2.00 kg
Net force acting on the object, F = (6.00 N) ihat - (7.70 N) jhat + (4.10 N) khat.
We can substitute these values into the equation F = m * a and solve for a.
F = m * a
(6.00 N) ihat - (7.70 N) jhat + (4.10 N) khat = (2.00 kg) * a
Comparing the coefficients of each unit vector, we can find the acceleration vector:
a = ((6.00 N) / (2.00 kg)) ihat + ((-7.70 N) / (2.00 kg)) jhat + ((4.10 N) / (2.00 kg)) khat
So, the acceleration of the object is:
a = (3.00 m/s^2) ihat + (-3.85 m/s^2) jhat + (2.05 m/s^2) khat.
(b) To find the angular momentum of the object about the origin, we can use the equation for angular momentum: L = r x p, where L is the angular momentum, r is the position vector, and p is the linear momentum.
Given:
Position vector, r = (2.00 m) ihat + (4.00 m) jhat - (3.00 m) khat
Linear momentum, p = m * v = (2.00 kg) * ((-6.20 m/s) ihat + (2.70 m/s) jhat + (2.60 m/s) khat)
To find the cross product of r and p, we can use the determinant of a 3x3 matrix:
L = |ihat jhat khat |
|2.00 4.00 -3.00 |
|-6.20 2.70 2.60 |
Expanding the determinant, we can calculate the value of the cross product.
L = (4.00 * 2.60 - 2.70 * (-3.00)) ihat - (2.00 * 2.60 - (-6.20) * (-3.00)) jhat + (2.00 * (-3.00) - 4.00 * (-6.20)) khat
Simplifying the expression, we get:
L = (26.2) ihat + (23.2) jhat + (8.40) khat
So, the angular momentum of the object about the origin is:
L = (26.2) ihat + (23.2) jhat + (8.40) khat.
(c) To find the torque about the origin acting on the object, we can use the equation for torque: τ = r x F, where τ is the torque, r is the position vector, and F is the force vector.
Given:
Position vector, r = (2.00 m) ihat + (4.00 m) jhat - (3.00 m) khat
Force vector, F = (6.00 N) ihat - (7.70 N) jhat + (4.10 N) khat
Using the cross product formula:
τ = r x F = |ihat jhat khat |
|2.00 4.00 -3.00 |
|6.00 -7.70 4.10 |
Expanding the determinant, we can calculate the value of the cross product.
τ = (4.00 * 4.10 - (-7.70) * (-3.00)) ihat - (2.00 * 4.10 - 6.00 * (-3.00)) jhat + (2.00 * (-7.70) - 4.00 * 6.00) khat
Simplifying the expression, we get:
τ = (43.7) ihat - (8.20) jhat + (-50.80) khat
So, the torque about the origin acting on the object is:
τ = (43.7) ihat - (8.20) jhat + (-50.80) khat.
(d) To find the angle between the velocity of the object and the force acting on the object, we can use the dot product formula:
θ = cos^(-1)((v • F) / (|v| * |F|)), where θ is the angle, v is the velocity vector, and F is the force vector.
Given:
Velocity vector, v = (-6.20 m/s) ihat + (2.70 m/s) jhat + (2.60 m/s) khat
Force vector, F = (6.00 N) ihat - (7.70 N) jhat + (4.10 N) khat
Using the dot product formula:
θ = cos^(-1)(((-6.20) * 6.00) + (2.70 * (-7.70)) + (2.60 * 4.10)) / (√((-6.20)^2 + 2.70^2 + 2.60^2) * √(6.00^2 + (-7.70)^2 + 4.10^2)))
Evaluating the dot product and the magnitudes, we get:
θ = cos^(-1)(-37.20 / (9.23 * 11.47))
Calculating the angle using a calculator, we find:
θ ≈ cos^(-1)(-0.319) ≈ 69.84°
So, the angle between the velocity of the object and the force acting on the object is approximately 69.84°.