How Do Forces Affect Motion in 2D Kinematics?

[rjs123]

Homework Statement

Lets say I have a .5kg mass traveling at 1200 m/s in the -x direction with 0 acceleration. The force applied to this mass is (2.3i + .093j)N. I have to find position, velocity, and acceleration at say 1 second after the force is applied. No gravitational force involved.

The Attempt at a Solution

F = ma will give me the acceleration. Should I just consider the force applied in the x direction? 2.3 N? I've only dealt with 1-d kinematics to this point...it seems the y force will only come into play for position...I could be wrong though.

 

[Freeze3018]

Vector approach is suitable for this kind of problem . resolve the velocities and forces and apply Newton's eqn of motion and equations of motion in straight line to solve the velocities and position at every time in x and y direction independently!

 

[rjs123]

Vector approach is suitable for this kind of problem . resolve the velocities and forces and apply Newton's eqn of motion and equations of motion in straight line to solve the velocities and position at every time in x and y direction independently!

thanks i see how to do it now.

 

[Freeze3018]

Ur welcome buddy!

 

[rwisz]

I would approach this problem by first identifying the given information and the unknowns. The given information includes the mass of the object (0.5kg), its initial velocity (1200 m/s in the -x direction), and the applied force (2.3i + .093j)N. The unknowns are the position, velocity, and acceleration of the object after 1 second.

To find the acceleration, we can use the equation F = ma, where F is the applied force and m is the mass of the object. Since the force is given in vector form, we can separate it into its x and y components. The x component of the force is 2.3N, which will result in an acceleration of 2.3 m/s^2 in the -x direction. The y component of the force is 0.093N, which will not contribute to the acceleration since there is no acceleration in the y direction.

Next, we can use the equations of motion to find the position and velocity of the object after 1 second. Since the object has an initial velocity and constant acceleration in the x direction, we can use the equation x = x0 + v0t + 1/2at^2 to find the position. Plugging in the values, we get x = 1200 m - 2.3 m/s^2 * 1 s^2 = 1197.7 m. This means that after 1 second, the object will be 1197.7 meters away from its initial position in the -x direction.

To find the velocity, we can use the equation v = v0 + at. Plugging in the values, we get v = 1200 m/s - 2.3 m/s^2 * 1 s = 1197.7 m/s. This means that after 1 second, the object will still be traveling at 1197.7 m/s in the -x direction.

In conclusion, the position, velocity, and acceleration of the object after 1 second can be determined by considering the applied force in the x direction and using the equations of motion. It is important to note that the y component of the force does not affect the motion of the object in this scenario. This problem showcases the application of 2-dimensional kinematics and how it can be used to solve real-world problems.