I think its A let me know its wrong or not
Answer:
hmax = 1/2 · v²/g
Explanation:
Hi there!
Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.
KE = PE
Where KE is the initial kinetic energy and PE is the final potential energy.
The kinetic energy of the ball is calculated as follows:
KE = 1/2 · m · v²
Where:
m = mass of the ball
v = velocity.
The potential energy is calculated as follows:
PE = m · g · h
Where:
m = mass of the ball.
g = acceleration due to gravity (known value: 9.81 m/s²).
h = height.
At the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:
PE = KE
m · g · hmax = 1/2 · m · v²
Solving for hmax:
hmax = 1/2 · v² / g
Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 ×
m
dynamic viscosity = 1.75 ×
Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ 
so
= µ 
............1
put here value
= 1.75× × 
= 0.175 v
and
area between air and puck is given by
Area = 
area = 
area = 7.85 × 
m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 × 
force = 1.374 × v
and now apply newton second law
force = mass × acceleration
- force = 
- 1.374 × v = 
t = 
time = 2.18
so time required after impact for a puck is 2.18 seconds
Work is calculated as the product of Force, Distance, and
angular motion. In this case, the work done by gravity is perpendicular to the motion
of the cart, so θ = 90°
and W=Fdcosθ
W=35.0 N x 20.0 m x cos90
W=0 J
This means that work done perpendicular to the direction of
the motion is always zero.
