Question

The initial velocity of a 4.0kg box is 11m/s due west. After the box slides 4.0m horizontally its speed is 1.5 m/s. Determine the magnitude and the direction of the non conservative force acting on the box as it slides.

Answer 1

Answer:

The magnitude and direction of the non conservative force acting on the box is <F= 59.36 N - DUE EAST DIRECTION>.

Explanation:

m= 4 kg

Vi= 11 m/s

Vf= 1.5 m/s

d= 4m

d= Vi * t - a * t²/2

clearing a:

a= 2*(Vi * t - d)/ t²

Vf= Vi - a * t

replacing "a" and clearing t:

t= 2d/(Vf+Vi)

t= 0.64 s

found now the value of a:

a= Vi - Vf / t

a= 14.84 m/s ²

F= m * a

F= 59.36 N