
A meter stick of mass \[400g\] is pivoted at one end and displaced through an angle \[60\] . The increase in its potential energy is
Answer
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Hint:We need to calculate the potential energy of the meter stick. We all know the potential energy is the multiplication of mass, gravity, and height from the ground. First, find the initial potential energy and final potential energy and subtract final potential energy from initial potential energy, then we can easily find the increase in potential energy.
Formula used:
The potential energy is represented as, \[V = mgl\]
Where, \[m\]- the mass of the meter stick
\[g\]- acceleration due to gravity,
\[l\]- length of displacement from the ground
Complete step by step solution:
The mass of the stick is \[m = 0.4kg\],
Length of the stick is \[l = 1m\],
The potential energy is written as, \[V = \dfrac{{mgl}}{2}\]
The center of the scale is displaced from the ground is half of the height, that’s why here we put \[\dfrac{l}{2}\]
The initial potential energy is \[{V_1} = \dfrac{{mgl}}{2}\]
Substitute the values, then we get the initial potential energy as, \[{V_1} = \dfrac{{0.4 \times 10 \times 1}}{2}\]
\[ \Rightarrow {V_1} = 2J\]
The final potential energy is, \[{V_2} = \dfrac{{mgl(1 - \cos \theta )}}{2}\]
The \[\theta = 60^\circ \], then,
\[{V_2} = \dfrac{{0.4 \times 10 \times 1(1 - 0.5)}}{2}\]
\[ \Rightarrow {V_2} = 1J\]
The increased potential energy is, \[V = {V_2} - {V_1}\]
\[V = 2 - 1 = 1J\]
The decreased potential energy is \[1J\]
Note: If the scale is pivoted at the angle \[180^\circ \] means it has maximum potential energy. Because the scale is displaced the maximum angle with respect to the initial position. After that, the angle is in between \[180^\circ - 360^\circ \] the potential energy is going to decrease. Because the displacement is in the opposite direction.
Formula used:
The potential energy is represented as, \[V = mgl\]
Where, \[m\]- the mass of the meter stick
\[g\]- acceleration due to gravity,
\[l\]- length of displacement from the ground
Complete step by step solution:
The mass of the stick is \[m = 0.4kg\],
Length of the stick is \[l = 1m\],
The potential energy is written as, \[V = \dfrac{{mgl}}{2}\]
The center of the scale is displaced from the ground is half of the height, that’s why here we put \[\dfrac{l}{2}\]
The initial potential energy is \[{V_1} = \dfrac{{mgl}}{2}\]
Substitute the values, then we get the initial potential energy as, \[{V_1} = \dfrac{{0.4 \times 10 \times 1}}{2}\]
\[ \Rightarrow {V_1} = 2J\]
The final potential energy is, \[{V_2} = \dfrac{{mgl(1 - \cos \theta )}}{2}\]
The \[\theta = 60^\circ \], then,
\[{V_2} = \dfrac{{0.4 \times 10 \times 1(1 - 0.5)}}{2}\]
\[ \Rightarrow {V_2} = 1J\]
The increased potential energy is, \[V = {V_2} - {V_1}\]
\[V = 2 - 1 = 1J\]
The decreased potential energy is \[1J\]
Note: If the scale is pivoted at the angle \[180^\circ \] means it has maximum potential energy. Because the scale is displaced the maximum angle with respect to the initial position. After that, the angle is in between \[180^\circ - 360^\circ \] the potential energy is going to decrease. Because the displacement is in the opposite direction.
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