The two fields were physical quantities are used in motion calculations are length and mass with time.
The physical quantity in a field is referred as every point in a particular space time.
<h3>
How physical quantities are used in motion calculations?</h3>
If we consider an object, the physical property of the object is considered as physical quantity and to measure that object is known as units. The Physical quantity can be classified as elemental physical quantity and derived physical quantity. Length, mass, time, etc.. are elemental physical quantity, momentum, density, acceleration, etc... are derived physical quantity. Only for charge and temperature the physical quantity will be less than zero.
Length, mass and time are the physical quantities used in motion calculations.
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Answer:
Kf= 36 J
W(net) = 32 J
Explanation:
Given that
m = 2 kg
F= 4 N
t= 2 s
Initial velocity ,u= 2 m/s
We know that rate of change of linear momentum is called force.
F= dP/dt
F.t = ΔP
ΔP = Pf - Pi
ΔP = m v - m u
v= Final velocity
By putting the values
4 x 2 = 2 ( v - 2)
8 = 2 ( v - 2)
4 = v - 2
v= 6 m/s
The final kinetic energy Kf
Kf= 1/2 m v²
Kf= 0.5 x 2 x 6²
Kf= 36 J
Initial kinetic energy Ki
Ki = 1/2 m u²
Ki= 0.5 x 2 x 2²
Ki = 4 J
We know that net work is equal to the change in kinetic energy
W(net) = Kf - Ki
W(net) = 36 - 4
W(net) = 32 J
Given data;
Fn = 10 N
Fs = 7 N
Fe = 4 N
force in X direction (Fx) = 4 N
force in Y direction (Fy) = 10-7 = 3 N
Net force (Fnet) = Sq.root[(Fx)² + (Fy)²]
= Sq root [ 4² + 3² ]
= 25 N
<em> Net force acting = 25 N</em>
Acceleration is given by:

where
v is the final velocity
u is the initial velocity
t is the time interval
Let's apply the formula to the different parts of the problem:
A) 
Let's convert the quantities into SI units first:


t = 4.0 min = 240 s
So the acceleration is

B) 
As before, let's convert the quantities into SI units first:


t = 94 s
So the acceleration is

C) 
For this part we have to use a different formula:

where we have
v = 0 is the final velocity
u = 89.2 m/s is the initial velocity
a is the acceleration
d = 75 m is the distance covered
Solving for a, we find
