a car of mass 750kg has a horizontal driving force of 2KN acting on it. its has a forward horizontal acceleration of 2ms-2. what is the resistive force acting horizontally ?
If possible I'd like to see the working for my better understanding
thanks in advance
If possible I'd like to see the working for my better understanding
thanks in advance
-
Given:
Acceleration of the car, a = 2 m/s^2
Applied force, F = 2 kN
mass of the car, m = 750 kg
Solution:
The problem can be illustrated as in the figure given below:
Apply the Newton's second law of motion to determine the net force on the car as follows:
F-f =ma
where
f - restricting force (Friction)
m - mass of the car
a - acceleration of the car
F - applied force
so,
2000 N - f = 750 kg × 2 m/s^2
f = 500 N
or
f = 0.5 kN
The negative sign indicates that the friction acts in the -ve x direction.
http://d.yimg.com/hd/answers/i/b403e89d457648509c8a821b06adac1a_A.jpeg?a=answers&mr=0&x=1383157944&s=1a8e353bf1a4d8b315290f6b528da29b
Acceleration of the car, a = 2 m/s^2
Applied force, F = 2 kN
mass of the car, m = 750 kg
Solution:
The problem can be illustrated as in the figure given below:
Apply the Newton's second law of motion to determine the net force on the car as follows:
F-f =ma
where
f - restricting force (Friction)
m - mass of the car
a - acceleration of the car
F - applied force
so,
2000 N - f = 750 kg × 2 m/s^2
f = 500 N
or
f = 0.5 kN
The negative sign indicates that the friction acts in the -ve x direction.
http://d.yimg.com/hd/answers/i/b403e89d457648509c8a821b06adac1a_A.jpeg?a=answers&mr=0&x=1383157944&s=1a8e353bf1a4d8b315290f6b528da29b
-
ma=F-Ff
then
Ff=F-ma
Ff=2x10^3-750*2
Ff=500 N
then
Ff=F-ma
Ff=2x10^3-750*2
Ff=500 N