If the mass of the crate is doubled but the initial velocity is not changed, the crate slides distance d before stopping.
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Let's recall the formula of Kinetic Energy as follows:
[tex]\large {\boxed {E_k = \frac{1}{2}mv^2 }[/tex]
Ek = Kinetic Energy ( Newton )
m = Object's Mass ( kg )
v = Speed of Object ( m/s )
Let us now tackle the problem !
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Given:
initial height = h₁ = 26 m
final height = h₂ = 16 m
initial speed = v₁ = 0 m/s
coefficient of friction = μ
gravitational acceleration = g
distance = d
Asked:
final speed = v₂ = ?
Solution:
We will use Work and Energy formula to solve this problem as follows:
[tex]W = \Delta Ek[/tex]
[tex]-f d = Ek_{final} - Ek_{initial}[/tex]
[tex]-\mu N d = \frac{1}{2}m (v_f)^2 - \frac{1}{2} m (v_i)^2[/tex]
[tex]-\mu mg d = \frac{1}{2}m (v_f)^2 - \frac{1}{2} m (v_i)^2[/tex]
[tex]-\mu mg d = \frac{1}{2}m (0)^2 - \frac{1}{2} m (v_i)^2[/tex]
[tex]-\mu mg d = \frac{1}{2} m (v_i)^2[/tex]
[tex]\mu g d = \frac{1}{2} (v_i)^2[/tex]
[tex]d = \frac{1}{2} (v_i)^2 \div ( \mu g )[/tex]
[tex]\boxed {d = \frac { (v_i)^2 } { 2 \mu g } }[/tex]
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From information above we can conclude that the distance is independent to the mass of the crate.
If the mass of the crate is doubled but the initial velocity is not changed, the crate slides the same distance d before stopping.
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Grade: High School
Subject: Physics
Chapter: Dynamics
