Two forces,
1 = (3.85 − 2.85) N
and
2 = (2.95 − 3.65) N,
act on a particle of mass 2.10 kg that is initially at rest at coordinates
(−2.30 m, −3.60 m).
(a) What are the components of the particle's velocity at t = 11.8 s?
= m/s
(b) In what direction is the particle moving at t = 11.8 s?
° counterclockwise from the +x-axis
(c) What displacement does the particle undergo during the first 11.8 s?
Δ = m
(d) What are the coordinates of the particle at t = 11.8 s?
x = m
y = m
(a) The components of the particle's velocity can be calculated using Newton's Second Law, F = ma. Since the particle is initially at rest, the sum of the forces acting on it will be equal to the mass multiplied by the acceleration. The components of the velocity can then be determined by dividing the sum of the forces by the mass and multiplying by the time (t = 11.8 s).
For force 1: F1 = (3.85 - 2.85) N = 1 N
For force 2: F2 = (2.95 - 3.65) N = -0.7 N
Sum of forces: F = F1 + F2 = 1 N + (-0.7 N) = 0.3 N
Using Newton's Second Law: F = ma, we have:
0.3 N = (2.10 kg) * a
a = 0.3 N / (2.10 kg) = 0.143 m/s^2
To find the components of the velocity, we multiply the acceleration by the time (t = 11.8 s):
vx = 0.143 m/s^2 * 11.8 s = 1.686 m/s (in the x-direction)
vy = 0 m/s (in the y-direction since the particle is initially at rest)
Therefore, the particle's velocity at t = 11.8 s is (1.686 m/s, 0 m/s).
(b) Since the particle has no velocity in the y-direction, it is only moving in the x-direction. Therefore, the particle is moving along the +x-axis.
(c) The displacement of the particle during the first 11.8 s can be calculated by finding the area under the velocity-time graph. Since the particle is initially at rest and only moving in the x-direction, the displacement is equal to the velocity in the x-direction multiplied by the time (t = 11.8 s):
Displacement = vx * t = 1.686 m/s * 11.8 s = 19.879 m
Therefore, the particle undergoes a displacement of 19.879 m during the first 11.8 s.
(d) To find the coordinates of the particle at t = 11.8 s, we start with the initial coordinates (-2.30 m, -3.60 m) and add the displacement calculated in part (c).
x_coordinate = initial_x_coordinate + Displacement = -2.30 m + 19.879 m = 17.579 m
y_coordinate = initial_y_coordinate + Displacement = -3.60 m + 0 m = -3.60 m
Therefore, the coordinates of the particle at t = 11.8 s are (17.579 m, -3.60 m).
To solve this problem, we will first calculate the net force acting on the particle using the given forces, and then use Newton's second law to find the acceleration. With the acceleration, we can use the kinematic equations to determine the velocity and displacement of the particle. Let's go step by step:
Step 1: Calculate the net force
To find the net force, we need to calculate the sum of the two given forces.
Net force = Force 1 + Force 2
Using the given forces:
1 = (3.85 - 2.85) N
2 = (2.95 - 3.65) N
Net force = (3.85 - 2.85) N + (2.95 - 3.65) N = 3.85 N - 2.85 N + 2.95 N - 3.65 N
Net force = 3.85 N - 2.85 N + 2.95 N - 3.65 N = 0 N
Since the net force is 0 N, we can conclude that the particle is in equilibrium.
Step 2: Calculate the acceleration
Since the particle is in equilibrium, the net force acting on it is zero. According to Newton's second law, the force acting on an object is equal to its mass multiplied by its acceleration.
Net force = mass * acceleration
Since the net force is zero, the acceleration must also be zero.
Step 3: Calculate the velocity
Since the acceleration is zero, the velocity of the particle remains constant.
Therefore, the components of the particle's velocity at t = 11.8 s are the same as at the beginning, which is 0 m/s.
(a) The components of the particle's velocity at t = 11.8 s are 0 m/s.
Step 4: Determine the direction of motion
Since the particle's velocity is zero, it is not moving in any direction.
Therefore, the direction of motion at t = 11.8 s is undefined.
(b) The particle is not moving at t = 11.8 s.
Step 5: Calculate the displacement
Since the particle is not moving, the displacement it undergoes during the first 11.8 s is 0 m.
(c) The displacement of the particle during the first 11.8 s is 0 m.
Step 6: Calculate the coordinates at t = 11.8 s
Since the particle is not moving, its coordinates will remain the same.
Given initial coordinates:
x = -2.30 m
y = -3.60 m
(d) The coordinates of the particle at t = 11.8 s are:
x = -2.30 m
y = -3.60 m
To find the components of the particle's velocity at t = 11.8 s and the direction of its movement, we need to apply Newton's second law of motion and use the given forces to calculate the acceleration. Then, we can use the kinematic equations to find the velocity components and determine the direction.
(a) To find the acceleration, we can use Newton's second law of motion:
F = ma
where F is the net force and m is the mass of the particle. In this case, the net force is the sum of the two given forces. So, we have:
F = F1 + F2
Substituting the values:
F = (3.85 − 2.85) N + (2.95 − 3.65) N
F = 1 N + (-0.7 N)
F = 0.3 N
Now, substituting the mass and the net force into Newton's second law:
a = F/m
a = 0.3 N / 2.10 kg
a ≈ 0.143 m/s²
Next, we can use the kinematic equations to find the velocity components. Since the particle is initially at rest (v₀ = 0), we have:
v = v₀ + at
Substituting the given values:
vx = 0 + (0.143 m/s²)(11.8 s)
vx ≈ 1.686 m/s
vy = 0 + (0.143 m/s²)(11.8 s)
vy ≈ 1.686 m/s
Therefore, the components of the particle's velocity at t = 11.8 s are approximately vx = 1.686 m/s and vy = 1.686 m/s.
(b) To determine the direction of the particle's movement at t = 11.8 s, we can use the formula:
θ = tan^(-1)(vy / vx)
Substituting the given values:
θ = tan^(-1)(1.686 m/s / 1.686 m/s)
θ ≈ 45°
Since the particle is moving counterclockwise from the +x-axis, the direction of movement at t = 11.8 s is 45° counterclockwise from the +x-axis.
(c) To find the displacement of the particle during the first 11.8 s, we can use the formula:
Δx = v₀x + 0.5 * a * t²
Δy = v₀y + 0.5 * a * t²
Given that the initial velocity is zero (v₀ = 0), the formulas simplify to:
Δx = 0.5 * a * t²
Δy = 0.5 * a * t²
Substituting the values:
Δx = 0.5 * 0.143 m/s² * (11.8 s)²
Δx ≈ 0.095 m
Δy = 0.5 * 0.143 m/s² * (11.8 s)²
Δy ≈ 0.095 m
Therefore, the displacement of the particle during the first 11.8 s is approximately Δx = 0.095 m in the x-direction and Δy = 0.095 m in the y-direction.
(d) To find the coordinates of the particle at t = 11.8 s, we can add the initial coordinates to the displacements found in part (c):
x = -2.30 m + 0.095 m
x ≈ -2.205 m
y = -3.60 m + 0.095 m
y ≈ -3.505 m
Therefore, the coordinates of the particle at t = 11.8 s are approximately x = -2.205 m and y = -3.505 m.
F1 = (x,y) = (3.85N,-2.85N)
F2 = (x,y) = (2.95N,-3.65N).
a. Fr=X+Yi=(3.85+2.95) + (-2.85i-3.65i)=
6.8 - 6.5i = 9.41N[316.3o] = Resultant
force.
2.10kg particle
Location: (x,y) = (-2.30m,-3.60m)
a(x) = Fx/mass = 6.8/2.1 = 3.24 m/s^2
Vx=Vox + a*t = 0 + 3.24*11.8= 38.21 m/s.
a(y) = Fy/mass = -6.5/2.1 = -3.10 m/s^2
Vy=Voy + a*t = 0 - 3.1*11.8= 36.52 m/s.
b. Tan A = Vy/Vx = 36.52/38.21=0.95587
A = 43.71o
c. d(x) = (Vx^2-Vox^2)/2a =
(38.21^2-0)/6.48 = 225.3 m.
Dx = 225.3 - (-2.30) = 227.6 m. = Hor.
comp. of displacement.
d(y) = (Vy^2-Voy^2)/2a =
(36.52^2-0)/-6.2 = -215.1 m.
Dy = -215.1 - (-3.60) = -211.5 m. = Ver.
comp. of displacement.
Tan Ar = Dy/Dx = -211.5/227.6 = -0.92933
Ar = -42.9o
Displacement=Dx/cosA=227.6/cos42.9=310.7
m.
X = 225.3 m
Y = -215.1 m.
d. (d(x),d(y)) = (225.3m,-215.1m)