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statuscvo [17]
2 years ago
13

An automobile with a mass of 1250 kg accelerates at a rate of 4 m/sec2 in the forward direction. What is the net force acting on

the automobile?​
Physics
1 answer:
kotegsom [21]2 years ago
6 0

\large{\underline{\underline{\maltese{\pink{\pmb{\sf{ \; Question \; :- }}}}}}}

An automobile with a mass of 1250 kg accelerates at a rate of 4 m/sec² in the forward direction. What is the net force acting on the automobile?

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

\large{\underline{\underline{\maltese{\green{\pmb{\sf{ \; Given\; :- }}}}}}}

  • ➢ Mass of the automobile = 1250 kg
  • ➢ Acceleration of the body = 4m/sec²

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

\large{\underline{\underline{\maltese{\orange{\pmb{\sf{ \; To \: Find \; :- }}}}}}}

  • Net force acting on the automobile = ?

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

\large{\underline{\underline{ \bigstar{{\pmb{\sf{ \; Concept \: and  \: Formula  \: Used \; :- }}}}}}}

➯ Force is defined as the product of the mass of the body and its acceleration.

\bigstar \:  \small{ \fbox {\textsf {\textbf{F = ma}}}}

<u>Where</u><u> </u><u>:</u>

  • F = Force
  • m = Mass of the body
  • a = acceleration due to gravity

\large{\underline{\underline{ \bigstar{{\pmb{\sf{ \; Substituting \: the\: Given\: Values  :- }}}}}}}

⇝ \:  \:  \small \textsf {\textbf{ F = (1250 × 4) N }}

⇝ \:  \:  \small \textsf {\textbf{ F = 5000 N }}

\large{\underline{\underline{ \bigstar{{\pmb{\sf{ \; Therefore \; :- }}}}}}}

❝ The net force acting on the automobile is 5000 N. ❞

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In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land
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Answer:

(a) \vec{J}=176.4\frac{kg*m}{s} \hat{j}

(b) F=2125.30N

Explanation:

(a) According to the law of conservation of energy, the potential energy of the person at 0.40 m is equal to its kinetic energy before the colision with the floor:

\Delta U=\Delta K\\mgh=\frac{mv^2}{2}\\v=\sqrt{2gh}\\v=\sqrt{2(9.8\frac{m}{s^2})(0.40m)}\\v=2.8\frac{m}{s}

This is the initial velocity in the negative y-direction. Impulse is given by:

\vec{J}=\Delta \vec{p}\\\vec{J}=m\vec{v_f}-m\vec{v_i}\\\vec{J}=63kg(0 \hat{j})-63kg(-2.8\frac{m}{s} \hat{j})\\\vec{J}=176.4\frac{kg*m}{s} \hat{j}

(b) The average force is:

F=\frac{J}{\Delta t}\\F=\frac{176.4\frac{kg*m}{s}}{0.083s}\\F=2125.30N

6 0
3 years ago
A 900-kg car cruising at a constant speed of 60 km/h is to accelerate to 100 km/h in 4 s. The additional power needed to achieve
kiruha [24]

To solve this problem we will apply the concepts related to power as a function of the change of energy with respect to time. But we will consider the energy in the body equivalent to kinetic energy. The change in said energy will be the difference between the two velocity data given by half of the mass. We will first convert the given units into an international system like this

Initial Velocity,

V_i = 60km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V_i = 16.6667m/s

Final Velocity,

V_f = 100km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V_f = 27.7778m/s

Now Power is defined as the change of Energy over the time,

P = \frac{E}{t}

But Energy is equal to Kinetic Energy,

P = \frac{\frac{1}{2} m\Delta v^2}{t}

P = \frac{\frac{1}{2} m(v_f^2-v_i^2)}{t}

Replacing,

P = \frac{\frac{1}{2} (900)(27.7778^2-16.6667^2)}{4}

P = 56kW

Therefore the correct answer is A.

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3 years ago
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