A 1700kg rhino charges at a speed of 50.0km/h. what is the magnitude of the average force needed to bring the rhino to a stop in 0.50s? express your answer to two significant figures.

First, let's convert the initial speed of the rhino:
[tex]v_i = 50 km/h = 13.9 m/s[/tex]
The acceleration of the rhino is given by:
[tex]a= \frac{v_f-v_i}{t}= \frac{0-13.9 m/s}{0.5 s}=-27.8m/s^2 [/tex]
where the negative sign means it is a deceleration. We can neglect the negative sign since we are interested only in the magnitude of the force applied, which is given by Netwon's second law:
[tex]F=ma = (1700 kg)(27.8 m/s^2)=4.7 \cdot 10^4 N[/tex]

johanrusli johanrusli · Answer 2 · 2019-12-14T04:20:49+01:00

The magnitude of the average force needed to bring the rhino to a stop in 0.50s is about 4.7 × 10⁴ Newton

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Further explanation

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

[tex]\large {\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

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Given:

mass of rhino = m = 1700 kg

initial speed of rhino = u = 50.0 km/h = 13⁸/₉ m/s

final speed of rhino = v = 0 m/s

time taken = t = 0.50 s

Asked:

average force = F = ?

Solution:

[tex]F = ma[/tex]

[tex]F = m \times \frac{v - u}{t}[/tex]

[tex]F = 1700 \times \frac{0 - 13\frac{8}{9}}{0.50}[/tex]

[tex]F = 1700 \times \frac{-250}{9}[/tex]

[tex]F \approx -4.7 \times 10^4 \texttt{ Newton}[/tex]

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Learn more

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Newton's Law of Motion: https://brainly.com/question/10431582
Example of Newton's Law: https://brainly.com/question/498822

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant