To tackle this problem, we need to analyze the forces acting on the particle and apply Newton's second law of motion. Let's break it down step by step.
Understanding the Forces
We have a particle with a mass of 5 kg that is initially at rest. It is acted upon by two forces: F1 and F2. The force F1 is given as 70 N in the positive y-direction (70j N). The force F2's direction and magnitude are unknown, but we know that the particle experiences a constant acceleration.
Applying Newton's Second Law
According to Newton's second law, the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F_net = m * a). The net force is the vector sum of all the forces acting on the particle.
- F_net = F1 + F2
- F1 = 70j N
- F2 = unknown
Since the particle is experiencing constant acceleration, we can express the net force as:
F_net = m * a
Finding the Missing Force F2
Let’s denote the acceleration vector as a. Since we don’t have its exact direction or magnitude yet, we can express it in terms of its components. For simplicity, let’s assume the acceleration has components a_x and a_y:
F_net = 5 kg * (a_x i + a_y j)
Now, substituting the forces into the equation:
F1 + F2 = 5 kg * (a_x i + a_y j)
We can express F2 as:
F2 = 5 kg * (a_x i + a_y j) - 70j N
To find F2, we need to know the acceleration. However, we can also express the net force in terms of its components:
- F_net_x = F2_x
- F_net_y = 70 N + F2_y
Acceleration Calculation
Since the particle starts from rest and experiences constant acceleration, we can use the kinematic equation:
s = ut + (1/2)at²
At t = 0, the initial velocity (u) is 0, so:
s = (1/2)at²
Assuming the particle moves in the y-direction under the influence of F1 and F2, we can express the displacement in the y-direction as:
s_y = (1/2)a_y t²
At t = 10 seconds, we can find the displacement in the y-direction:
s_y = (1/2)a_y (10)² = 50a_y
Velocity Calculation
The velocity of the particle at any time t can be calculated using the equation:
v = u + at
Since the initial velocity is zero, we have:
v_y = 0 + a_y * t
At t = 10 seconds:
v_y = a_y * 10
Final Steps
To find the exact values of F2 and the velocity vector, we need to know the acceleration a_y. If we assume the particle is only acted upon by F1 in the y-direction, we can find the acceleration due to F1:
Using F = ma:
70 N = 5 kg * a_y
Thus, a_y = 14 m/s². Now we can substitute this back into our equations:
F2 = 5 kg * (0 i + 14 j) - 70j N = 70j N - 70j N = 0 N
So, F2 is zero. The particle only experiences the force from F1.
The velocity vector at t = 10 seconds is:
v_y = 14 m/s² * 10 s = 140 m/s
Therefore, the velocity vector of the particle at t = 10 seconds is:
v = 0 i + 140 j m/s
In summary, the missing force F2 is 0 N, and the velocity vector of the particle at t = 10 seconds is 140 j m/s.
