Respuesta :
Answer:
a. -312.4 N or 312.4 N [E] b. -312.4 N or 312.4 N [E]
Explanation:
a. Using v² = u² + 2as where u = initial velocity of car = 64.8 km/h = 64.8 × 1000 m/3600 s = 18 m/s, v = final velocity of car = 0 m/s since the car stops, a = deceleration of car and s = distance car moved = 729 m.
So, making the acceleration subject of the formula, we have
a = (v² - u²)/2s
substituting the values of the variables, we have
a = ((0 m/s)² - (18 m/s)²)/(2 × 729 m)
= - (18 m²/s²)/(2 × 729 m)
= -324 m²/s² ÷ 1458 m
= -0.22 m/s²
So, the net force , F acting on the car is F = ma where m = mass of car = 1420 kg and a = deceleration = -0.22 m/s².
So, F = ma = 1420 kg × -0.22 m/s² = 312.4 N or 312.4 N [E]
b. Since no forward force acts on the car while it is slowing down, the frictional force equals the force acting to stop the car, f = ma which is equal to the net force where m = mass of car = 1420 kg and a = deceleration = -0.22 m/s²
So, f = ma = 1420 kg × -0.22 m/s² = -312.4 N or 312.4 N [E]
a) The net force acting on the car will be 312.4 N [E]
(b) The force of friction acting on the car  will be 312 N [E].
What is the friction force?
It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically it is defined as the product of the coefficient of friction and normal reaction.
The given data in the problem is;
m is the mass of car=1420 kg
u is the initial speed 64.8 km/h = 4.8 × 1000 m/3600 s = 18 m/s
v is the final speed =0 m/sec
s is the distance travelled before stop= 729 m
a is the deceleration = -0.22 m/s².
a) The net force acting on the car will be  312.4 N [E]
The value of net force is found with the help of Newtons third equation of motion;
[tex]\rm v^2=u^2+2as\\\\ \rm a= \frac{v^2-u^2}{2s} \\\\ \rm a= \frac{(0)^2-(18)^2}{2\times 729} \\\\ \rm a= 0.22 \ m/s^2[/tex]
So the net force will be;
[tex]\rm F = ma \\\\\ \rm F = 1420 \times 0.22 ² \\\\ \rm F =312.4\ N [E][/tex]
Hence the net force acting on the car will be  312.4 N [E]
(b) The force of friction acting on the car  will be 316 N [E].
The frictional force equals the force working to stop the automobile,
f = ma,
Which is equal to the net force,
The force of friction will be
[tex]\rm F_f= m \times a \\\\\ \rm F_f= 1420\times -0.22 \\\\ \rm F_f= -312.4 N [E][/tex]
Hence the force of friction acting on the car  will be 312 N [E].
To learn more about the friction force refer to the link;
https://brainly.com/question/1714663
