Besides the gravitational force a 325kg object is subjected

Besides the gravitational force, a 3.25-kg object is subjected to one other constant force. The object starts from rest and in 1.20 s experiences a displacement of (5.25i 3.30j) m, where the direction of j is the upward vertical direction. Determine the other force.

Solution

the force is defined as follows
               F = fx i + fy j ;

gravitational force = - 3.25 * 9.8 j = -31.85 j ;

for x axis ;

displacement = 5.25 and time = 1.2 s ;

so, the distance is,

           s = (1/2) a t ^2

5.25 = (1/2)*a * 1.2^2 ;

therefore, x-component of acceleratrion:

   a_x = 7.29 m /s ^2

hence, the force in x-axis:

                fx = m ax = 3.25* 7.29 = 23.6925 N = 23.7 N

in y-direction:

      s = 0.5 a t^2 ;

- 3.3 = 0.5 * a * 1.2^2 ;

     a = -4.583 m /s ^2 ;

thus, the force in y-axis is,

      fy - mg = m ay ;

fy = m ( g + ay )

                 = 3.25 ( 9.8 - 4.583 )

=16.95525 N

                = 16.96 N

Therefore, the required force: F = [23.7 i + 16.96 j] N