time : 4 sec ,mass : 7000 kg, velocity : 36 km / hr. find the force applied to Stop the bus.( answer is - 17500N ) I will follow you if you can answer this questions
Answers
Answer:
Mass of the bus = 7000 kg
Time = 4 sec
Velocity of the bus = 36 km/hr
To Find :
The Force applied to stop the bus
Solution :
using first equation of motion , v = u + at
Here ,
v is final velocity
u is initial velocity
a is acceleration
t is time
We have ,
v = 0 {since it comes to rest}
u = 36 km/hr = 36(5/18) m/s = 10 m/s
t = 4 sec
Mass = 7000 kg
Substituting the values ;
➙ 0 = 10 + (F/m)(4) { ∵ a = F/m}
➙ 0 = 10 + (F/7000)(4)
➙ 0 = 10 + (4F/7000)
➙ 0 = (70000 + 4F)/7000
➙ 0 = 70000 + 4F
➙ -70000 = 4F
➙ F = -70000/4
➙ F = -17500 N \begin{gathered}\\\end{gathered}
Hence ,
The Force that should be applied to bring the bus to rest is -17,500
Answer:
Given data:
Mass of stone(m)=7000 kg
Distance(S)=400 meter
Time(t)=20 second
1) We have to find the acceleration of the truck
So from the equation of motion S=ut+21a×t2, whereu=0
S=21a×t2
a=t22×S=20×202×400=2sec2m
2) Force acting on truck
So,from newton's 2nd law of motion
F=ma=7000×2=14000N=14kN