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  • A force causes an acceleration (Newton’s second law $\mathbf{F}=m\mathbf{a}$).
  • Next, acceleration is by definition the rate of change of velocity $\mathbf{a}:=\frac{d\mathbf{v}}{dt}$.
  • Speed is by definition the magnitude of velocity $v=|\mathbf{v}|=\sqrt{\mathbf{v}\cdot\mathbf{v}}$.

So, the first bullet point tells us that if there is a non-zero force then there is a non-zero acceleration, and so by the second bullet point, there will be a change in the velocity. But, does this mean the speed changes? No, not necessarily. To figure out what causes a change in the speed, and how this is related to velocity and acceleration, let’s do some basic calculus: \begin{align} \frac{dv}{dt}&=\frac{1}{2\sqrt{\mathbf{v}\cdot\mathbf{v}}}\cdot 2\mathbf{v}\cdot\frac{d\mathbf{v}}{dt}=\frac{1}{v}(\mathbf{v}\cdot \mathbf{a}). \end{align} So, if we want the speed to change, i.e if we want $\frac{dv}{dt}$ to be non-zero, then we need the dot product $\mathbf{v}\cdot\mathbf{a}$ to be non-zero, i.e we need the acceleration and velocity vectors of the particle to be non-zero, and we also need their directions to be such that the component of the acceleration along the direction of the velocity is non-zero.

It is tradition to call the direction of the velocity vector $\mathbf{v}$ the tangential direction (because the velocity is by definition the tangent vector to the curve traced out by the particle). So, putting all this together, a non-zero force always causes a change in velocity, but in order for it to change the speed, we need the direction of the force to have a non-zero tangential component (i.e a component along the velocity), hence the answer to your question

Does that mean Kinetic energy will change only when there is a resultant force component that is parallel to the velocity component?

is yes.

  • A force causes an acceleration (Newton’s second law $\mathbf{F}=m\mathbf{a}$).
  • Next, acceleration is by definition the rate of change of velocity $\mathbf{a}:=\frac{d\mathbf{v}}{dt}$.
  • Speed is by definition the magnitude of velocity $v=|\mathbf{v}|=\sqrt{\mathbf{v}\cdot\mathbf{v}}$.

So, the first bullet point tells us that if there is a non-zero force then there is a non-zero acceleration, and so by the second bullet point, there will be a change in the velocity. But, does this mean the speed changes? No, not necessarily. To figure out what causes a change in the speed, and how this is related to velocity and acceleration, let’s do some basic calculus: \begin{align} \frac{dv}{dt}&=\frac{1}{2\sqrt{\mathbf{v}\cdot\mathbf{v}}}\cdot 2\mathbf{v}\cdot\frac{d\mathbf{v}}{dt}=\frac{1}{v}(\mathbf{v}\cdot \mathbf{a}). \end{align} So, if we want the speed to change, i.e if we want $\frac{dv}{dt}$ to be non-zero, then we need the dot product $\mathbf{v}\cdot\mathbf{a}$ to be non-zero, i.e we need the acceleration and velocity vectors of the particle to be non-zero, and we also need their directions to be such that the component of the acceleration along the direction of the velocity is non-zero.

It is tradition to call the direction of the velocity vector $\mathbf{v}$ the tangential direction (because the velocity is by definition the tangent vector to the curve traced out by the particle). So, putting all this together, a non-zero force always causes a change in velocity, but in order for it to change the speed, we need the direction of the force to have a non-zero tangential component (i.e a component along the velocity).

  • A force causes an acceleration (Newton’s second law $\mathbf{F}=m\mathbf{a}$).
  • Next, acceleration is by definition the rate of change of velocity $\mathbf{a}:=\frac{d\mathbf{v}}{dt}$.
  • Speed is by definition the magnitude of velocity $v=|\mathbf{v}|=\sqrt{\mathbf{v}\cdot\mathbf{v}}$.

So, the first bullet point tells us that if there is a non-zero force then there is a non-zero acceleration, and so by the second bullet point, there will be a change in the velocity. But, does this mean the speed changes? No, not necessarily. To figure out what causes a change in the speed, and how this is related to velocity and acceleration, let’s do some basic calculus: \begin{align} \frac{dv}{dt}&=\frac{1}{2\sqrt{\mathbf{v}\cdot\mathbf{v}}}\cdot 2\mathbf{v}\cdot\frac{d\mathbf{v}}{dt}=\frac{1}{v}(\mathbf{v}\cdot \mathbf{a}). \end{align} So, if we want the speed to change, i.e if we want $\frac{dv}{dt}$ to be non-zero, then we need the dot product $\mathbf{v}\cdot\mathbf{a}$ to be non-zero, i.e we need the acceleration and velocity vectors of the particle to be non-zero, and we also need their directions to be such that the component of the acceleration along the direction of the velocity is non-zero.

It is tradition to call the direction of the velocity vector $\mathbf{v}$ the tangential direction (because the velocity is by definition the tangent vector to the curve traced out by the particle). So, putting all this together, a non-zero force always causes a change in velocity, but in order for it to change the speed, we need the direction of the force to have a non-zero tangential component (i.e a component along the velocity), hence the answer to your question

Does that mean Kinetic energy will change only when there is a resultant force component that is parallel to the velocity component?

is yes.

Source Link
peek-a-boo
peek-a-boo
  • 6.7k
  • 1
  • 11
  • 27

  • A force causes an acceleration (Newton’s second law $\mathbf{F}=m\mathbf{a}$).
  • Next, acceleration is by definition the rate of change of velocity $\mathbf{a}:=\frac{d\mathbf{v}}{dt}$.
  • Speed is by definition the magnitude of velocity $v=|\mathbf{v}|=\sqrt{\mathbf{v}\cdot\mathbf{v}}$.

So, the first bullet point tells us that if there is a non-zero force then there is a non-zero acceleration, and so by the second bullet point, there will be a change in the velocity. But, does this mean the speed changes? No, not necessarily. To figure out what causes a change in the speed, and how this is related to velocity and acceleration, let’s do some basic calculus: \begin{align} \frac{dv}{dt}&=\frac{1}{2\sqrt{\mathbf{v}\cdot\mathbf{v}}}\cdot 2\mathbf{v}\cdot\frac{d\mathbf{v}}{dt}=\frac{1}{v}(\mathbf{v}\cdot \mathbf{a}). \end{align} So, if we want the speed to change, i.e if we want $\frac{dv}{dt}$ to be non-zero, then we need the dot product $\mathbf{v}\cdot\mathbf{a}$ to be non-zero, i.e we need the acceleration and velocity vectors of the particle to be non-zero, and we also need their directions to be such that the component of the acceleration along the direction of the velocity is non-zero.

It is tradition to call the direction of the velocity vector $\mathbf{v}$ the tangential direction (because the velocity is by definition the tangent vector to the curve traced out by the particle). So, putting all this together, a non-zero force always causes a change in velocity, but in order for it to change the speed, we need the direction of the force to have a non-zero tangential component (i.e a component along the velocity).