Explanation:
It is given that,
Mass of the stone, m = 20 kg
Frictional force, F = -2 N
A player releases a stone that travels 27.9 m before coming to rest, s = 27.9 m
We need to find the initial velocity of the stone. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2as[/tex]
v = 0
And, [tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{-2\ N}{20\ kg}=-0.1\ m/s^2[/tex]
[tex]0-u^2=2\times -0.1\ m/s^2\times 27.9\ m[/tex]
[tex]u=2.36\ m/s[/tex]
So, the speed of the stone when it was released is 2.36 m/s. Hence, this is the required solution.
The speed of the stone when the player released it was about 2.4 m/s
[tex]\texttt{ }[/tex]
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 !
[tex]\texttt{ }[/tex]
Given:
mass of stone = m = 20 kg
distance traveled = d = 27.9 m
magnitude of friction force = f = 2.0 N
final speed of stone = v = 0 m/s
Asked:
initial speed of stone = u = ?
Solution:
Firstly, we will use Newton's Second Law of Motion to calculate the deceleration of the stone:
[tex]\Sigma F = ma[/tex]
[tex]-f = m a[/tex]
[tex]-2.0 = 20 a[/tex]
[tex]a = -2.0 \div 20[/tex]
[tex]\boxed{a = -0.1 \texttt{ m/s}^2}[/tex]
[tex]\texttt{ }[/tex]
Next, we could calculate the initial speed of stone as follows:
[tex]v^2 = u^2 + 2ad[/tex]
[tex]0^2 = u^2 + 2( -0.1) (27.9)[/tex]
[tex]u^2 = 5.58[/tex]
[tex]u = \sqrt{5.58}[/tex]
[tex]\boxed{u \approx 2.4 \texttt{ m/s}}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Dynamics
