Answer:
Velocity [tex]V = \frac{dx}{dt} = \frac{d}{dt} (P + Qt + Rt^{3})[/tex]
[tex]V = 0 + Q(1) + 3Rt^{2}[/tex]
V = Q + 3Rt²
at t = 0, [tex]v_{i}[/tex] = Q + 3r(0) ==> Q
at t = T, [tex]v_{f}[/tex] = Q + 3rT²
Work done (W) = ΔKE = [tex]\frac{1}{2} m(v_f^{2} - v_i^{2})[/tex]
W = [tex]\frac{1}{2} m[(Q + 3RT^{2})^{2} - Q^{2}][/tex]
W = [tex]\frac{1}{2} m[/tex] [Q² + 9R²T⁴ + 2Q(3RT²) - Q²]
W = [tex]\frac{1}{2} m[/tex] (9R²T⁴ + 6QRT²)
Explanation:
Differentiate the position.
Find the equation for speed.
Find the initial and final speed.
Use work energy theorem to find the work done by finding the change in kinetic energy.
The work done by the force from t = 0 to T can be expressed by the equation [tex]\ \dfrac 12 m (9R^T^2 + 6QRT^2)[/tex].
Work can be defined as the product of the applied force and displacement of an object.
Differentiate the position,
[tex]V = \dfrac {dx }{dt}\\V = \dfrac d {dt }(P+Qt+Rt^3)\\V = 0 + Q\times 1 + 3 Rt^2[/tex]
The initial speed at [tex]T= 0[/tex], the initial velocity [tex]v_i = Q[/tex]
The final speed at t = T. the final velocity [tex]v_f = Q +3 rT^2[/tex]
From work- energy formula,
[tex]W = \dfrac 12 m (v_f ^2 - v_i^2 )[/tex]
Put the values in the formula,
[tex]W = \dfrac 12 m (Q +3RT^2 )^2- Q^2\\\\W = \dfrac 12 m [Q² + 9R^2T^4+ 2Q(3RT^2) - Q^2]\\\\\ W = \dfrac 12 m (9R^T^2 + 6QRT^2)[/tex]
Therefore, the work done by the force from t = 0 to T can be expressed by the equation [tex]\ \dfrac 12 m (9R^T^2 + 6QRT^2)[/tex].
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