A 6.71 kg object has the following acceleration components:
ax = ( 0.51 m/s2) + ( 0.75 m/s3) t
ay = ( 11.9 m/s2) + ( 0.59 m/s3) t
A) What is the magnitude of the net force acting on the object at time t = 6.07 s? In Newtons: ______
B) What is the direction of the net force at this same time? Give your answer as a number of degrees counter-clockwise from the x axis. In Degrees: _________
You have to add vectors, taking the angle between them into account.
At t= 6.07 s
a_x = ( 0.51 m/s2) + ( 0.75 m/s3) *6.07= 5.0625 m/s�
a_y = ( 11.9 m/s2) + ( 0.59 m/s3) *6.07= 15.48 m/s�
You can add these components and use F=ma, where 'a' is the resultant acceleration. or you can work out Fx and Fy and combine these. It doesn't' make any difference....
F_x = m*a_x = 6.71 x 5.0625 = 33.96N
F_y = m*a_y = 6.71 x 15.48= 103.87 N
|F| = ?(F_x� + F_y�)
= ?(33.96� + 103.87�)
Angle = tan-1(F_y/F_x) = tan-1(103.87/33.98) = 71.88?