Question

3) A mass of 9 kg is at rest on a horizontal surface. One force of 40 N is applied at an angle of 18 degrees below the horizontal (in the +x direction), and a force of 9 N is applied vertically upward (perpendicular and away from the surface). If these forces are applied for 2.3 seconds, what is the velocity (in the x direction) of the mass in m/s? If in the -x direction, include a negative sign.

Answer 1

By applying Newton's 2nd law

Fnet,x =m*ax

ax =Fnet,x/(m)

ax =(40 N*cos 18)/(9 kg)

ax =4.2269 m/s^2

From kinematic equation

v=u+a*t

at rest ,u =0 m/s

v =4.2269 m/s^2*2.3 s

v =9.720 m/s