To find the magnitude of the \( x \)-component of force acting on the particle at point \( P(1, 2, 3) \), we can use the concept of potential energy and force relationship. The force components are given by the negative gradient of the potential energy function.
The potential energy function is given as:
\(U = (2x^2 + 3y^3 + 2z)\)
The force acting on the particle is given by:
\(\vec{F} = -\nabla U\)
Here, \(\nabla U\) represents the gradient of the potential energy function \(U\). The gradient in three dimensions is:
\(\nabla U = \left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z} \right)\)
Let's calculate each partial derivative:
Partial derivative with respect to \(x\):
\(\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(2x^2) = 4x\)
Partial derivative with respect to \(y\):
\(\frac{\partial U}{\partial y} = \frac{\partial}{\partial y}(3y^3) = 9y^2\)
Partial derivative with respect to \(z\):
\(\frac{\partial U}{\partial z} = \frac{\partial}{\partial z}(2z) = 2\)
Substituting these partial derivatives, the force is:
\(\vec{F} = - (4x, 9y^2, 2)\)
To find the \(x\)-component of the force at \(P(1, 2, 3)\), substitute \(x = 1\), \(y = 2\), and \(z = 3\):
\(F_x = -4x = -4(1) = -4 \, \text{N}\)
The magnitude of the \(x\)-component of force is:
\(|F_x| = 4 \, \text{N}\)
Thus, the magnitude of the \(x\)-component of the force acting on the particle at point \(P(1, 2, 3)\) is 4 N.
\[ U = 2x^2 + 3y^3 + 2z \]
The force in the \(x\)-direction is given by \(F_x = -\frac{\partial U}{\partial x}\). Differentiating \(U\) with respect to \(x\):
\[ F_x = -\frac{\partial}{\partial x}(2x^2) = -4x \]
Substitute \(x = 1\):
\[ F_x = -4 \times 1 = -4 \]
The magnitude of \(F_x\) is 4 N.
So, the correct answer is: 4 N
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
