Related Formula
Instantaneous power P$P$ generated by a force is given by:
P = F · v$$P = \vec{F} \cdot \vec{v}$$
Newton's second law:
a = Fm = d vdt v = ∫ a dt$$\vec{a} = \frac{\vec{F}}{m} = \frac{\mathrm{d}\vec{v}}{\mathrm{d}t} \implies \vec{v} = \int \vec{a} \,\mathrm{d}t$$
Core Logic
Convert mass to SI units:
m = 1000 ~g = 1 ~kg$$m = 1000 \mathrm{~g} = 1 \mathrm{~kg}$$
Calculate acceleration:
a = 2t i + 3t² j1 = 2t i + 3t² j$$\vec{a} = \frac{2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}}{1} = 2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}$$
Step 1: Determine Velocity Vector
Assuming the object starts from rest at t = 0$t = 0$:
v = ∫₀t (2t i + 3t² j) dt = t² i + t³ j$$\vec{v} = \int_{0}^{t} (2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}) \,\mathrm{d}t = t^2\hat{\mathbf{i}} + t^3\hat{\mathbf{j}}$$
Step 2: Calculate Power
Compute the dot product of force and velocity:
P = F · v = (2t i + 3t² j) · (t² i + t³ j)$$P = \vec{F} \cdot \vec{v} = (2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}) \cdot (t^2\hat{\mathbf{i}} + t^3\hat{\mathbf{j}})$$
P = (2t)(t²) + (3t²)(t³) = 2t³ + 3t⁵ ~W$$P = (2t)(t^2) + (3t^2)(t^3) = 2t^3 + 3t^5 \mathrm{~W}$$Pattern Recognition
Sees: Time-dependent force
$
Pattern Recognition
Sees: Time-dependent force $
\vec{F} \propto t^n
on a 1 kg mass. Shortcut: For$ on a 1 kg mass.
Shortcut: For $m=1
kg, velocity is the integral of the force components. Power is the dot product of the force vector and its integral. Since$ kg, velocity is the integral of the force components. Power is the dot product of the force vector and its integral. Since $\int at^n \mathrm{d}t = \frac{a}{n+1} t^{n+1}
, power component becomes$, power component becomes $\frac{a^2}{n+1} t^{2n+1}
. Here,$. Here, $2^2/2 t^3 + 3^2/3 t^5 = 2t^3 + 3t^5 \mathrm{~W}$.
Chapter Mix
Class 11 Physics: Work, Energy and Power
