Question

If a 1.00 kg body has an acceleration of 3.00 m/s^2 at 67° to the positive direction of the x axis, then what are (a) the x component and (b) the y component of the net force on it, and (c) what is the net force in unit-vector notation?

Answer 1

(a) Fx= 1.17 N

(b) Fy = 2.76 N

(c) F = 1.17 N i^ + 2.76j^

Step-by-step explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data

m= 1 kg

a= 3.00 m/s² at 67° to the positive direction of the x axis

x-y components of the acceleration

ax= a* cos 67° = 3*cos67°= 1,17 m/s²

ax= a* sin 67° = 3*sin67°= 2.76 m/s²

Problem development

We apply the formula (1)

(a)∑Fx = m*ax

Fx= 1 kg* 1,17 m/s² = 1.17 N

(b)∑Fy = m*ay

Fy = 1 kg* 2.76 m/s² = 2.76 N

(c) F = 1.17 N i^ + 2.76j^