Answer:
Part a)

Part b)

Explanation:
As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same
So we will have




Part b)
By equation of kinetic energy we have




Answer:
No, not necessarily
Explanation:
If an object is moving with an acceleration that causes its speed to be reduced, there will be a moment in which it reaches v = 0, but this doesn't necessarily mean that the acceleration isn't acting anymore. If the object continues its movement with the same acceleration, it's velocity will become negative.
An example of an object that has zero velocity but non-zero acceleration:
If you throw an object in the air with a certain velocity, it will move vertically, reducing its velocity in a 9,8
rate (which is the acceleration caused by gravity). At a certain point, the object will reach its maximum height, and will start to fall. In the exact moment that it reaches the maximum height, before it starts falling, its velocity is zero, but gravity is still acting on the object (this is the reason why it starts falling instead of just being stopped at that point). Therefore, at that point, the object has zero velocity but an acceleration of 9,8
.
Explanation:
The first equation of motion in kinematics is given by :
.....(1)
u is initial speed
a is acceleration
v is final speed
t is time
Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.
The horizontal force needed to start the calculator moving from rest is 1.5 N
What is Kinetic friction?
It is defined as a force that acts between moving surfaces.
The magnitude of the force will depend on the coefficient of kinetic friction between the two materials.
Here,
weight of calculator, N = 3 N
The coefficients of static frictions, µ (static) = 0.50
The coefficients of kinetic frictions, µ (kinetic) = 0.40
Now,
The horizontal force required = The static friction force
F = µ (static) * weight of calculator
F = 0.50 * 3.0
F = 1.5 N
Hence,
The horizontal force needed to start the calculator moving from rest is 1.5 N
Learn more about horizontal force here:
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Can we see the diagram? Thanks.