Mechanics (Part – 3)

Sitting on one side of the beam, under the influence of the force of gravity, this part bends down towards the earth and the other side rises up.

It acts like a swing for kids.

• v=u+at
  

v = final velocity

u= initial velocity

a= acceleration due to gravity

t = time elapsed (1 second in this case)

For a freely falling object, the initial velocity (u) is 0 because the object starts from rest.

The equation becomes:
                                                                  $$v=0+g\cdot t$$

Substituting the values:
v=9.8 m/s²×1 second
v=9.8m/s

The velocity of the object when it collides with the ground will be 9.8 m/s downward.

To calculate the work done, we can use the formula:

• Work Done=F×d×cos⁡(θ)
  

F is the force applied (200 N),

$d$ is the displacement (4 m),

θ is the angle between the force and the direction of displacement.

If the force is applied in the same direction as the displacement (i.e.,θ=0°  the cosine of 0° is 1,)
so the formula simplifies to:
                                                                                          Work Done=F×d

Now, substituting the values:

                                                                      Work Done=200 N×4 m = 800

The work done is 800 joules (J)

                                                                                 a=Δv/Δt ​
Where:

Δv is the change in velocity (final velocity - initial velocity)

Δt is the time taken for the change in velocity

Given:

• Initial velocity, vi = 25 m/s
  
• Final velocity, vf = 30 m/s
  
• Time, Δt=10 seconds
  

Now, calculate the change in velocity:Δv=vf​−vi​=30m/s−25m/s=5m/s

Now, substitute into the acceleration formula:= (5 m/s) / (10 seconds) = 0.5 m/s²

• First, let us convert kilometer to meters and hours to seconds, individually.
• We know that 1 kilometer = 1000 meters and 1 hour = 3600 seconds.

Now, 1 km/h can also be written as 1000/3600: 1 (kilometer/hour) = 1000/3600 (meters/seconds).

So, this is the formula which is used to convert km/h to m/s.

On simplifying 1000/3600 further, we get 5/18.

• At this point, the object momentarily stops before it starts to fall back down due to gravity.
• The velocity becomes zero at the maximum height because all the upward motion is counteracted by the force of gravity, and the object is about to change direction.

An object is in static equilibrium when it is at rest and the net force acting on it is zero. This means:

• The object has no acceleration.
• The vector sum of all the forces acting on the object is zero.
• The object is not moving or rotating.

In other words, the forces acting on the object balance each other out, resulting in no change in its motion.

This can be expressed as:
                                                                                  F=ma
Where:

• F is the force applied on the object,
• m is the mass of the object,
• a is the acceleration of the object.

From this equation, we can derive:
1. Acceleration is directly proportional to force: If the mass is constant, increasing the applied force will increase the acceleration.
2. Acceleration is inversely proportional to mass: If the force is constant, increasing the mass of the object will decrease the acceleration.
In summary:

• a∝F (if mass is constant)
• a∝1/m ​(if force is constant)

So the acceleration remains constant when the ball is thrown up.

• The projectile moves under the influence of gravity: Gravity constantly acts downward, pulling the projectile towards the Earth's surface.
• The horizontal velocity remains constant (in ideal conditions): In the absence of air resistance, the horizontal component of the projectile's velocity does not change during its flight.
• The vertical velocity changes: As the projectile rises, the vertical component of its velocity decreases due to gravity. At the highest point (maximum height), the vertical velocity becomes zero. Afterward, the vertical velocity increases as the projectile falls back down.
• The trajectory is parabolic: The path followed by the projectile is typically a curved trajectory (a parabola) due to the combination of constant horizontal velocity and changing vertical velocity.
  Air resistance (if considered) affects the motion: In real-world conditions, air resistance slows down the horizontal velocity and affects the overall motion, but in idealized projectile motion (without air resistance), this is not considered.

In summary, during projectile motion:

• The horizontal velocity remains constant (ignoring air resistance).
• The vertical velocity changes due to gravity.
• The trajectory is parabolic.
• The motion is influenced by gravity throughout.