Answer: 29.50 m
Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:
f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).
f=m.a=μ*N= m*a= μ*m*g= m*a
then
a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2
With this value we can determine the short distance to stop the train
as follows:
x= vo*t- (a/2)* t^2
Vf=0= vo-a*t then t=vo/a
Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m
Answer:
159.1 ton
Explanation:
The solution is shown in the attached file
Answer:
Time = 80.91 seconds
Explanation:
Given the following data;
Velocity = 5.50 m/s.
Distance = 445 meters
To find the time;
Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.
Mathematically, velocity is given by the equation;

Substituting into the formula, we have;
5.5 = 445/time
Time = 445/5.5
Time = 80.91 seconds
Answer:
3.258 m/s
Explanation:
k = Spring constant = 263 N/m (Assumed, as it is not given)
x = Displacement of spring = 0.7 m (Assumed, as it is not given)
= Coefficient of friction = 0.4
Energy stored in spring is given by

As the energy in the system is conserved we have

The speed of the 8 kg block just before collision is 3.258 m/s