Question

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind is blowing. If the wind's force on the rope is negligible, what drag force in Newtons does the wind exert on the ball? B)A box is sliding down an incline tilted at a 15° angle above horizontal. The box is initially sliding down the incline at a speed of 1.4 m/s. The coefficient of kinetic friction between the box and the incline is 0.37. How far does the box slide down the incline before coming to rest?C)An object weighing 3.9 N falls from rest subject to a frictional drag force given by Fdrag = bv2, where v is the speed of the object and b = 2.5 N ∙ s2/m2. What terminal speed will this object approach?

Answer 1

Part a)

`F_v = 4.28 N`

Part B)

`L = 1.02 m`

Part C)

`v = 1.25 m/s`

Step-by-step explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

`Tcos\theta = mg`

`T sin\theta = F_v`

`(F_v)/(mg) = tan\theta`

`F_v = mg tan\theta`

`F_v = 1.2* 9.81 (tan20)`

`F_v = 4.28 N`

Part B)

Here we can use energy theorem to find the distance that it will move

`-\mu mg cos\theta L + mg sin\theta L = -(1)/(2)mv^2`

`(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - (1)/(2)m(1.4)^2`

`(-3.5 + 2.54)L = - 0.98`

`L = 1.02 m`

Part C)

At terminal speed condition we know that

`F_v = mg`

`bv^2 = mg`

`2.5 v^2 = 3.9`

`v = 1.25 m/s`