
A conservative force \[\vec F = (6.0x - 12)\hat{i}\,N\] where, $x$ is in metres, acts on a particle moving along an X axis. The potential energy $U$ associated with this force is assigned a value of $27J$ at $x = 0$ .
(a) Write an expression for U as a function of $x$ ,with U in Joules and $x$ in metres.
(b) What is the maximum positive potential energy? At what (c) negative value and (d) positive value of $x$ is the potential energy equal to zero?
Answer
409.5k+ views
Hint: In physics, when the force acts on a body and it causes a displacement in the body , it’s called body has done some work and when this work done by a force is independent of path taken, then it’s called conservative force and this conservative force is related with potential energy as $F = - \dfrac{{\partial U}}{{\partial x}}$.
Complete step by step answer:
(a) It’s given us that $U(0) = 27J$ and we can write $F = - \dfrac{{\partial U}}{{\partial x}}$ as
\[ - U = \int\limits_0^x {(6x - 12)dx} \]
$\Rightarrow - U = (3{x^2} - 12x)$
$\therefore U = 27 + 12x - 3{x^2}$
Hence, the function of U is $U(x) = 27 + 12x - 3{x^2}$
(b) To find maximum potential energy its derivative must be zero which is the given force, hence
\[(6.0x - 12) = 0\]
$x = 2$
Hence finding $U(x) = 27 + 12x - 3{x^2}$ at $x = 2$
$U(2) = 27 + 24 - 12$
${U_{\max }}(2) = 39Joule$
Hence, maximum potential energy is ${U_{\max }}(2) = 39\,Joule$
(c) Equating this equation $U(x) = 27 + 12x - 3{x^2}$ to zero we get,
$27 + 12x - 3{x^2} = 0$
Or Taking common factor we can write
$(x + 1.6)(x - 5.6) = 0$
Equate both factors to zero we get,
$\therefore x = - 1.6\,m$
(d) From part (c) we get, $(x + 1.6)(x - 5.6) = 0$
$x = 5.6$
Hence, $x = 5.6$ is the positive value of $x$ at which potential energy is zero.
Note: It should be remembered that, partial derivative of potential energy is taken because force is a vector quantity and its derivative has to be taken in every component’s direction. The relation between force and potential energy can also be written in form of gradient at $F = - \vec \nabla U$
Complete step by step answer:
(a) It’s given us that $U(0) = 27J$ and we can write $F = - \dfrac{{\partial U}}{{\partial x}}$ as
\[ - U = \int\limits_0^x {(6x - 12)dx} \]
$\Rightarrow - U = (3{x^2} - 12x)$
$\therefore U = 27 + 12x - 3{x^2}$
Hence, the function of U is $U(x) = 27 + 12x - 3{x^2}$
(b) To find maximum potential energy its derivative must be zero which is the given force, hence
\[(6.0x - 12) = 0\]
$x = 2$
Hence finding $U(x) = 27 + 12x - 3{x^2}$ at $x = 2$
$U(2) = 27 + 24 - 12$
${U_{\max }}(2) = 39Joule$
Hence, maximum potential energy is ${U_{\max }}(2) = 39\,Joule$
(c) Equating this equation $U(x) = 27 + 12x - 3{x^2}$ to zero we get,
$27 + 12x - 3{x^2} = 0$
Or Taking common factor we can write
$(x + 1.6)(x - 5.6) = 0$
Equate both factors to zero we get,
$\therefore x = - 1.6\,m$
(d) From part (c) we get, $(x + 1.6)(x - 5.6) = 0$
$x = 5.6$
Hence, $x = 5.6$ is the positive value of $x$ at which potential energy is zero.
Note: It should be remembered that, partial derivative of potential energy is taken because force is a vector quantity and its derivative has to be taken in every component’s direction. The relation between force and potential energy can also be written in form of gradient at $F = - \vec \nabla U$
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
