At the instant the displacement of a 2.00 kg object relative to the origin is d = (2.00 m) + (4.00 m) - (3.00 m) , its velocity is v = - (5.50 m/s) + (2.90 m/s) + (3.20 m/s) , and it is subject to a force F = (5.70 N) - (7.80 N) + (3.80 N) .
(a) Find the acceleration of the object.
( 1 m/s2) + ( 2 m/s2) + ( 3 m/s2)
(b) Find the angular momentum of the object about the origin.
( 4 kg · m2/s) + ( 5 kg · m2/s) + ( 6 kg · m2/s)
(c) Find the torque about the origin acting on the object.
( 7 N · m) + ( 8 N · m) + ( 9 N · m)
(d) Find the angle between the velocity of the object and the force acting on the object.
10°
To solve this problem, we'll use some basic concepts from physics. Let's break down each part of the question one by one.
(a) Find the acceleration of the object:
Acceleration is the rate at which velocity changes. We can calculate it using Newton's second law of motion, which states that force equals mass times acceleration (F = ma). Since the force acting on the object is given, we can use it to find the acceleration.
Given:
Force, F = (5.70 N) - (7.80 N) + (3.80 N) = 1.70 N
Mass, m = 2.00 kg
Using Newton's second law, we have:
F = ma
1.70 N = 2.00 kg * a
Solving for acceleration, we get:
a = 1.70 N / 2.00 kg = 0.85 m/s^2
Therefore, the acceleration of the object is 0.85 m/s^2.
(b) Find the angular momentum of the object about the origin:
Angular momentum is a vector quantity defined as the cross product of an object's position vector and its linear momentum. In this case, we need to find the angular momentum of the object about the origin.
Given:
Displacement, d = (2.00 m) + (4.00 m) - (3.00 m) = 3.00 m
Velocity, v = -(5.50 m/s) + (2.90 m/s) + (3.20 m/s) = 0.60 m/s
The angular momentum is calculated as:
L = r x p
Where r is the displacement vector and p is the linear momentum vector.
Using the magnitudes of r and p:
r = |d| = 3.00 m
p = |m * v| = |2.00 kg * 0.60 m/s| = 1.20 kg·m/s
L = r * p * sin(θ)
Since we are given the angle between the velocity and the force in part (d), we can use the same angle for θ in this part. Therefore, θ = 10°.
L = 3.00 m * 1.20 kg·m/s * sin(10°) = 0.628 kg·m^2/s
Therefore, the angular momentum of the object about the origin is approximately 0.628 kg·m^2/s.
(c) Find the torque about the origin acting on the object:
Torque is the rotational equivalent of force. It is calculated as the cross product of the position vector and the force vector. In this case, we need to find the torque about the origin.
Given:
Force, F = (5.70 N) - (7.80 N) + (3.80 N) = 1.70 N
Displacement, d = (2.00 m) + (4.00 m) - (3.00 m) = 3.00 m
The torque is calculated as:
τ = r x F
Using the magnitudes of r and F:
r = |d| = 3.00 m
F = |F| = |1.70 N| = 1.70 N
τ = r * F * sin(θ)
However, the angle θ is not given in this part. It is only given in part (d), so we cannot calculate the torque without this information.
(d) Find the angle between the velocity of the object and the force acting on the object:
Given:
Velocity, v = -(5.50 m/s) + (2.90 m/s) + (3.20 m/s) = 0.60 m/s
Force, F = (5.70 N) - (7.80 N) + (3.80 N) = 1.70 N
θ = arccos((v · F) / (|v| * |F|))
θ = arccos((0.60 m/s * 1.70 N) / (|0.60 m/s| * |1.70 N|))
θ = arccos(1.02 N·m / (1.02 N·m))
θ = arccos(1)
θ = 0°
Therefore, the angle between the velocity of the object and the force acting on the object is 0°.