S
swingoutk
Guest
A 1000-kg car moving north at 100km/h brakes to a stop in 50 m. What are the magnitude and direction of the force?
O
(Ohm)Mistress Bekki
Guest
For an object accelerating to rest:
vi^2 = 2ax
So your acceleration is:
a = vi^2 / 2x
So per Newton's 2nd Law, your force is:
F = ma = m vi^2 / 2x
The direction of the force is opposite the direction of the initial speed.
vi^2 = 2ax
So your acceleration is:
a = vi^2 / 2x
So per Newton's 2nd Law, your force is:
F = ma = m vi^2 / 2x
The direction of the force is opposite the direction of the initial speed.
R
Roonal.18™
Guest
from 2nd law of motion F=ma
m= 1000
a = ?
100 km/h = 27.7778 m/s
therefore
Vinitial = 27.7778 m/s
Vfinal = 0 m/s since car is stopping
From eq. v² = u² + 2as
where s is the distance moved = 50 m
v = Vfinal
u = Vinitial
0 = 27.778² +100(a)
a = -7.7162 m/s²
therefore magnitude of the force = |1000*-7.7162|
= 7716.2 N
= 7720 N (3.s.f)
Direction of the force is opposite to that of motion, since car is decelerating
m= 1000
a = ?
100 km/h = 27.7778 m/s
therefore
Vinitial = 27.7778 m/s
Vfinal = 0 m/s since car is stopping
From eq. v² = u² + 2as
where s is the distance moved = 50 m
v = Vfinal
u = Vinitial
0 = 27.778² +100(a)
a = -7.7162 m/s²
therefore magnitude of the force = |1000*-7.7162|
= 7716.2 N
= 7720 N (3.s.f)
Direction of the force is opposite to that of motion, since car is decelerating