AS
work done =W = F.d = F d cosФ (Ф is angle between force F and displacement d) If a body/object is moving on a smooth surface (friction-less surface ) .There is no force acting on that body. F=0 so W=FdcosФ= (0)dcosФ ⇒ W=0
Now if a body is facing some amount of force but under the action of force there is no displacement covered. d=0 so W =FdcosФ= F(0)cosФ ⇒W=0
example: A person is applying a force on rigid wall but wall remains at rest there is no displacement occurs in wall.
The third term upon which work done dependent is angle between force and displacement i.e Ф. If Ф=90° then W= FdcosФ= Fdcos90⇒ W=0 ( as cos 90°=0)
Answer:
128.9 N
Explanation:
The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:
where
F is the force
is the change in momentum
is the time interval
The change in momentum can be written as
where
m = 0.04593 kg is the mass of the ball
u = 0 is the initial velocity of the ball
is the final velocity of the ball
Substituting into the original equation, we find the force exerted on the golf ball:
Answer:
2361.6N
Explanation:
Mass of player = 82kg
Velocity = 1.2m/s
Kinetic energy of player:
= 1/2mv²
= 1/2*82*1.2²
= 41x1.44
= 59.04J
Final kinetic energy = 0
Change in kinetic energy
|∆k| = |0-59.04|
= 59.04
Workdone by the feet = fd
d = 0.025
Fd = 59.04
F = 59.04/0.025
= 2361.6N
This is his average force.
Explanation:
We define force as the product of mass and acceleration.
F = ma
It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.
Given Data:
Width of the pool = w = 50 ft
length of the pool = l= 100 ft
Depth of the shallow end = h(s) = 4 ft
Depth of the deep end = h(d) = 10 ft.
weight density = ρg = 62.5 lb/ft
Solution:
a) Force on a shallow end:
b) Force on deep end:
c) Force on one of the sides:
As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.
1) Force on the Rectangular part:
2) Force on the triangular part:
here
h = h(d) - h(s)
h = 10-4
h = 6ft
now add both of these forces,
F = 25000lb + 150000lb
F = 175000lb
d) Force on the bottom:
