What is the Unknown Force Acting on a 2.8 kg Object in Upward Vertical Motion?

[tdusffx]

can anyone help me start this problem?

Besides its weight, a 2.8 kg object is subjected to one other constant force. The object starts from rest and in 1.2s experieces a displacement of (4.2i - 3.3j) m, where the direction of j is the upward vertical direction. determine the other force

 

[Dick]

Start by splitting the displacement into i and j components. Using the displacements calculate the total force on the object. Then subtract the weight force from the total force to get the 'other force'.

 

[tdusffx]

kk, I got...

deltaX = Vixt + 1/2at^2

4.2 = 0 1/2a(1.2)^2

a = 5.8i

deltaY = Viyt + 1/2at^2

-3.3 = 0 + 1/2a(1.2)^2

Y = -5.4j

F = 2.8kg(5.8i + 5.4j) ?

 

[Dick]

Seems roughly ok. I don't have to check your arithmetic, right? But aren't you losing signs again in the total force? Shouldn't the y component of the force be DOWN?

 

[tdusffx]

yup yup! you're right again, dick. Signs are always messing me up. Thanks again.