Answer:
uk = 0.25
Explanation:
Given:-
- An object comes to stop with acceleration, a = -2.45 m/s^2
Find:-
What is the coefficient of kinetic friction between the object and the floor?
Solution:-
- Assuming the object has mass (m) that slides over a rough surface with coefficient of kinetic friction (uk). There is only Frictional force (Ff) acting in the horizontal axis on the object opposing the motion (-x direction).
- We will apply equilibrium equation on the object in vertical direction.
N - m*g = 0
N = m*g
Where, N : Contact force exerted by the surface on the floor
g : Gravitational acceleration constant = 9.81 m/s^2
- Now apply Newton's second law of motion in the horizontal ( x-direction ):
- Ff = m*a
- The frictional force is related to contact force (N) by the following expression:
Ff = uk*N
- Substitute the 1st and 3rd expressions in the 2nd equation:
uk*m*g = -m*a
uk = a / g
- Plug in the values and solve for uk:
uk = - (-2.45) / 9.81
uk = 0.25
Answer:
v = 0.42m/s
Explanation:
In order o calculate the linear speed of the point at the border of the wheel, you first take into account that the total acceleration of such a point is given by:
(1)
atotal: total acceleration = 1.2m/s^2
ar: radial acceleration of the wheel
at: tangential acceleration
The tangential acceleration is also given by:
(2)
r: radius of the wheel = (40cm/2 )= 20cm = 0.2m
α: angular acceleration = 4.0rad/s^2
You replace the expression (2) into the expression (1) and solve for the radial acceleration:
Next, you use the following formula for the radial acceleration and solve for the linear speed:
The linear speed of the point at the border of the wheel is 0.42m/s
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
Answer:
v = 30.39 m/s
Explanation:
given,
mass of glider,M = 680 Kg
mass of the skydiver, m = 68 Kg
horizontal velocity,V = 30 m/s
when skydiver releases, velocity,v = 30 m/s
velocity if the glider,v' = ?
use the conservation of momentum
M V = m' v' + m v
m' = 680-68 = 612 Kg
680 x 30 = 612 x v + 60 x 30
612 v = 18600
v = 30.39 m/s
since the skydiver's speed will be the same as before release
The glider should continue to travel at 30.39 m/s since there are no external forces acting on it
