A0.145 kg baseball pitched at 39.0 m/s is hit on a horizontal line drive straight back toward the pitcher at 52.0 m/s. if the contqct time bewteen bat and ball is 3.00x10^-3 calculate the average force between the bat and ball during contact

consider the velocity towards the pitcher as positive

m = mass of the baseball = 0.145 kg

v₀ = initial velocity of the baseball = - 39 m/s

v = final velocity of the baseball = 52 m/s

t = time of contact = 3 x 10⁻³ sec

F = average force between bat and ball

Using impulse-change in momentum equation

F t = m (v - v₀ )

F (3 x 10⁻³) = (0.145) (52 - (- 39))

F = 4398.33 N

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