Aparticle travels to the right at a constant rate of 6.8 m/s. it suddenly is given a vertical acceleration of 3.4 m/s 2 for 4.3 s. what is its direction of travel after the acceleration with respect to the horizontal? answer between −180◦ and +180◦ .

65.05°

Explanation:

Hello!

In order to know the direction of travel after the acceleration we must first know the velocity after the acceleration.

The horizontal component of velocity remains the same, but the vertical component will be given as the product of the (constant) acceleration times the total time.

Vx = 6.8 m/s

Vy = (3.4)*(4.3) m/s = 14.62 m/s

Then the angle of the motion of the particle with respect to the horizontal is given by:

Ф = arctan(Vy/Vx) =arctan(14.62/6.8) = 65.05°

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards