Answer to Question #82249 in Physics for Dema Ghazi

Question

Given the forces: F1= (10)x + (20)y , F2= (-7)x + (15)y and F3 (-20)y . applied on a particle of mass 40g .
a. Calculate the magnitude and the direction of the acceleration.
b. The particle is now split inoto two particles of equal masses, for the second mass the force F2=0 . What’s the acceleration of the second particle?

Answer 1

The Newton’s second law states

〖ma=F〗_net

F_net=F_1+F_2+F_3

=(10) x ̂ + (20) y ̂+(-7) x ̂ + (15) y ̂+(-20) y ̂=(3) x ̂+(15)y ̂

So, the acceleration of the crate

a_x=(F_net )_x/m =(3 N)/(0.040 kg)=75 m/s^2

a_y=(F_net )_y/m =(15 N)/(0.040 kg)=375 m/s^2

The magnitude of the acceleration

a=√(a_x^2+a_y^2 )=√(〖75〗^2+〖375〗^2 )=382 m/s^2

Direction

θ=tan^(-1)⁡〖a_y/a_x =tan^(-1)⁡〖375/75=〗 〗 79° to the x+direction

F_net=F_1+F_3

=(10) x ̂ + (20) y ̂+(-20) y ̂=(10) x ̂

〖a=a〗_x=(F_net )_x/m =(10 N)/(0.020 kg)=500 m/s^2