We know that the Delta E + W(Work done by non-conservative
forces) = 0 (change of energy)
In here, the non-conservative force is the friction force
where f = uN (u =kinetic friction coefficient)
W= f x d = uNd ; N=mg
Delta E = 1/2 mV^2 -1/2mVi^2
umgd + 1/2mV^2 - 1/2mVi^2 = 0 (cancel out the m term)
This will then give us:
1/2Vi^2-ugd = 1/2V^2
V^2 = Vi^2 - 2ugd
So plugging in our values, will give us:
V= Sqrt (5.6^2 -2.3^2)
=sqrt (26.07)
= 5.11 m/s
Answer:
I'm pretty sure its 3m/s^2 for the acceleration but I don't know the force part sorry .
Explanation:
15m/s - 0m/s divided by 5 s = 3m/s
I'm no expert or anything so I could be wrong but this is the best I can give you. Sorry
It is a wedge because of the ramp thingy
Answer:
-414.96 N
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
The force the ground exerts on the parachutist is -414.96 N
If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase