Respuesta :
Answer:
Net force on the car will be [tex]3689236.111N[/tex]
Explanation:
We have given mass of the car m = 850 kg
Initial velocity of the car u = 45 km/hr [tex]=45\times \frac{5}{18}=12.5m/sec[/tex]
As the car finally stops so final velocity of car v = 0 m/sec
Distance travel for stooping of car s = 1.8 cm = 0.018 m[tex]v^2=u^2+2as[/tex]
From third equation of motion [tex]v^2=u^2+2as[/tex]
So [tex]0^2=12.5^2+2\times a\times 0.018[/tex]
[tex]a=-4340.277m/sec^2[/tex]
So magnitude if acceleration will be [tex]a=4340.277m/sec^2[/tex]
So net force on the car [tex]F=ma=850\times 4340.277=3689236.111N[/tex]
So net force on the car will be [tex]3689236.111N[/tex]
Magnitude of the net force needed to stop in the distance equal to the diameter of a dime of 1.8 cm is 3.7 × 10⁶ N
Net force based problem:
What information do we have?
Mass of car = 850 kg
Speed of car = 45 km/h = 12.5 m/s
Using third equation of motion,
v² = u² + 2as
0² = 12.5² + 2(a)(0.018)
Acceleration = - 4340.27m/s²
F(net) = ma
F(net) = (850)(- 4340.27)
F(net) = - 3,689,236.7N
F(net) = 3.7 × 10⁶ N
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