A motor car of mass 800kg travelling at 20m/s is brought to rest by brake in 100m.calculate the average braking force required.in physics
Vf^2=vi^2 + 2ad
solve for acceleration a.
Then, remember Force=m*a ? solve for force.
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bobpursley
Feb 7, 2016
Vf^2 = Vi +2ad
Data
Vi=20.0 m/s
Vf=0m/s ( at rest)
dx=100m
a=?
Force?
(O)^2=(20)^2 +2a(100)
-400=200a
a=-2m/s^2
Force=ma
:= -2(800)
= -1600N
Vf^2 = Vi +2ad
Data
Vi=20.0 m/s
Vf=0m/s ( at rest)
dx=100m
a=?
Force?
(O)^2=(20)^2 +2a(100)
-400=200a
a=-2m/s^2
Force=ma
:= -2(800)
= -1600N
Vf^2=vi^2 + 2ad
solve for acceleration a.
Then, remember Force=m*a ? solve for force.
To calculate the average braking force required, we can use the principles of kinematics and Newton's second law of motion.
Step 1: Determine the initial velocity (u), final velocity (v), and distance (s) given in the problem.
- Initial velocity (u) = 20 m/s
- Final velocity (v) = 0 m/s (brought to rest)
- Distance (s) = 100 m
Step 2: Calculate the acceleration (a) using the kinematic equation:
v^2 = u^2 + 2as
Rearrange the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
Plugging in the values:
a = (0 - 20^2) / (2 * 100)
a = -400 / 200
a = -2 m/s²
The negative sign indicates deceleration, as the car is coming to a stop.
Step 3: Calculate the braking force (F) using Newton's second law of motion formula:
F = ma
Plugging in the values:
F = 800 kg * (-2 m/s²)
F = -1600 N (since the force is in the opposite direction to motion)
Therefore, the average braking force required to bring the car to rest is 1600 Newtons, and it acts in the opposite direction to the car's motion.