Mass.
Because mass doesn't depend on weight but weight depends on mass.
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The gravitational potential energy will increase by 423.36 J
<h3>How to determine the potential energy at ground level</h3>
- Mass (m) = 72 kg
- Acceleration due to gravity (g) = 9.8 m/s²
- Height (h) = 0 m
- Potential energy at ground level (PE₁) =?
PE = mgh
PE₁ = 72 × 9.8 × 0
PE₁ = 0 J
<h3>How to determine the potential energy at 60 cm (0.6 m)</h3>
- Mass (m) = 72 kg
- Acceleration due to gravity (g) = 9.8 m/s²
- Height (h) = 0.6 m
- Potential energy at 60 cm (0.6 m) (PE₂) =?
PE = mgh
PE₂ = 72 × 9.8 × 0.6
PE₂= 423.36 J
<h3>How to determine the change in potential energy </h3>
- Potential energy at ground level (PE₁) = 0 J
- Potential energy at 60 cm (0.6 m) (PE₂) = 423.36 J
- Change in potential energy =?
Change in potential energy = PE₂ - PE₁
Change in potential energy = 423.36 - 0
Change in potential energy = 423.36 J
Learn more about energy:
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Answer:
The linear velocity is 
Explanation:
According to the law of conservation of energy
The potential energy possessed by the hoop at the top of the inclined plane is converted to the kinetic energy at the foot of the inclined plane
The kinetic energy can be mathematically represented as

Where
is the moment of inertia possessed by the hoop which is mathematically represented as
Here R is the radius of the hoop
is the angular velocity which the hoop has at the bottom of the lower part of the inclined plane which is mathematically represented as

Where v linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface
Now expressing the above statement mathematically


=>
=> 
=> 
=> 
Substituting values


Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
(1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
(2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
![a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}](https://tex.z-dn.net/?f=a%5B%281%29e%5E%7B-6t%7D-6te%5E%7B-6t%7D%5D%3D0%5C%5C%5C%5C1-6t%3D0%5C%5C%5C%5Ct%3D%5Cfrac%7B1%7D%7B6%7D)
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):

hence, the maximum speed is v_max = ((1/6)e^-1)a