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mars1129 [50]
3 years ago
9

A car of mass 1200 kg traveling westward at 30. m/s is slowed to a stop in a distance of 50. m by the car’s brakes. What was the

braking force? Select one:
A. 1,800,000 N B. 0.80 N C. 11,000 N D. 2000 N
Physics
2 answers:
natka813 [3]3 years ago
5 0
I believe that the answer is C. 11,000
soldi70 [24.7K]3 years ago
5 0

Answer:

C. 11000 N

Explanation:

car kinematics :The car moves with uniformly accelerated movement for this movement the following formula apply:

vf²= v₀² + 2* a*d  Formula (1)

vf : final speed (m/s)

v₀ : initial speed (m/s)

a  : acceleration (m/s²)

d  : distance (m)

Car kinetics:  We apply Newton's second law:

∑F= m*a

∑F : Algebraic sum of forces (N)

m : mass (kg)

a  : acceleration (m/s²)

Acceleration Calculation

We replace Vf=0 , v₀= 30 m/s , d=50m  the  formula (1)

0= 30²+ 2*a*50

a= - 30² /2*50 = - 900/100 = -9 m/s²

Braking force Calculation

∑F= m*a

-F =  1200 kg*(-9 m/s²)

F = 10800N

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part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
  • mass of the larger disk = M_2\ =\ 1.6\ kg
  • Radius of the larger disk =R_2\ =\ 5.0\ cm\ =\ 0.05\ m
  • mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding,\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

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A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)

Net torque on the smaller disk,

\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)

From eqn (1) and (2), we get,

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let 'a_2' be the acceleration of the block in the second case,

From the above expression,

\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2

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