A23 kg suitcase is being pulled with a constant speed by a handle that is at an angle of 29° above the horizontal. if the normal force exerted on the suitcase is 160 n, what is the force f applied to the handle?

329.89N

Explanation:

Normal force is a force that is perpendicular to the horizontal surface. The normal force will therefore acts in the positive y direction.

To get the force F applied at the handle, we will resolve the force to the component of the normal force (y component)

The component of the force along the positive y axis (Fy) will be +Fsin(theta) where theta is 29°

Since both the normal force and the resolved force are acting in the same direction, Fy = normal force

Fsin(theta) = 160N

Fsin29°= 160N

F = 160/sin29°

F = 160/0.485

F = 329.89N

Therefore the force F applied to the handle is 329.89N

a=d/t

a="triangle"v/t

explanation:

a = acceleration