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
The answer I think is C -7.6 m/s
For the motion of the plane it is given to us that it travel 15.0 km in 15.3 s
So here we can use the formula of average velocity to find it
formula is given as ratio of total displacement and total time
here it is given that
time = 15.3 s
we need to convert the time into hour as the speed is required in km/h
so we know that
now by the above formula
so the velocity of the plane would be 3529.4 km/h
