Answer:
The angular velocity of Ball A will be greater than the angular velocity of Ball B when they reach the top of the hill.
Explanation:
Angular velocity can be defined as how fast an object rotates relative to a given point or frame of reference.
The question said the hill encountered by Ball A is frictionless, so Ball A will continue to rotate at the same rate it started with even when it reached the top of the hill.
Ball B on the other hand rolls without slipping over its hill, i.e there's friction to slow down its rotational motion which thus reduces how fast Ball B will rotate at the top of the hill
We can first calculate the net force using the given information.
By Newton's second law, F(net) = ma:
F(net) = 25 * 4.3 = 107.5
We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):
f = F(net) - F(app) = 150 - 107.5 = 42.5 N
Now we can calculate the coefficient of friction, u, using the normal force, F(N):
f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
First of all, we can find the mass of the person, since we know his weight W:

And so

We know for Newton's second law that the resultant of the forces acting on the person must be equal to the product between the mass and the acceleration a of the person itself:

There are only two forces acting on the person: his weight W (downward) and the vincular reaction Rv of the floor against the body (upward). So we can rewrite the previous equation as

We know the acceleration of the system,

(upward, so with same sign of Rv), so we can solve to find the value of Rv, the normal force exerted by the elevator's floor on the person:
Answer:
625000 N/ m
Explanation:
m= 20 kg
v= 30 m/s
x= 12 cm
k = ?
Here when the mass when hits at spring its speed is
Vi= 30 m/s
Finally it comes to rest after compressing for 12 cm
i-e Vf = 0 m/s
Distance= S= 12 cm = 0.12 m
using
2aS= Vf2 - Vi2
==> 2a ×0.12 = o- 30 × 30
==> a = 900 ÷ 0.24 = 3750 m/sec2
Now we know;
F = ma
F= -Kx
==> ma= -kx
==> 20 × 3750 = -K × 0.12
==> k = 625000 N/ m
Answer:
a) p = m1 v1 + m2 v2
, b) dp / dt = m1 a1 + m2 a2
, c) It is equivalent to force
dp / dt = 0
Explanation:
In this problem we have two blocks and the system is formed by the two bodies.
Part A. Initially they ask us to find the moment of the whole system
p = m1 v1 + m2 v2
Part B.
Find the derivative
dp / dt = m1 dv1dt + m2 dv2 / dt
dp / dt = m1 a1 + m2 a2
Part C.
Let's analyze the dimensions
m a = [kg] [m / s2] = [N]
It is equivalent to force
Part d
Acceleration is due to a net force applied
Part e
The acceleration of block 1 is due to the force exerted by block 2 during the moment change
Part f
Force of block 1 on block 2
True f12 = m1a1 f21 = m2a2
Part g
By the law of action and reaction are equal magnitude F12 = f21
Part H
dp / dt = 0
Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed