Question

A model airplane has momentum given by p=[(-0.75kg.m/s3)t2 + (3.0kg.m/s)] i + (0.25kg.m/s2)t j. Find the components Fx, Fy, and Fz of the net force on the airplane.

Answer 1

Answer:

$F_x$ = -1.5t

$F_y$ = 0.25

$F_{z}$ = 0

Explanation:

Given equation;

p = [(-0.75 kgm/s³)t² + (3.0 kgm/s)] i + (0.25 kgm/s²)t j.

From Newton's law, the rate of change of momentum of a body is the net force acting on that body. i.e

∑F = $\frac{dp}{dt}$       -----------(i)

Substitute the equation of p into equation (i) and differentiate with respect to t as follows;

∑F = $\frac{dp}{dt}$ = $\frac{d| [(-0.75)t^{2} + (3.0)] i + (0.25)t j|}{dt}$

∑F = $\frac{dp}{dt}$ = $[-1.5t + 0]i + 0.25j$

∑F = $[-1.5t + 0]i + 0.25j$

But

∑F = $F_xi + F_yj + F_zk$

Where;

$F_x, F_y, F_z$ are the components of the net force in the x, y and z direction respectively.

=> $F_xi + F_yj + F_zk$ =  $[-1.5t + 0]i + 0.25j$ = $-1.5ti + 0.25j$

=> $F_x$ = -1.5t

=> $F_y$ = 0.25

=> $F_{z}$ = 0