A)A 1.1-kg ball is hanging from the end of a rope. The rope hangs at an angle 26

Question

A)A 1.1-kg ball is hanging from the end of a rope. The rope hangs at an angle 26° from the vertical when a 10 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? (Give you answer to the nearest 0.01 N)

B)A box is sliding down an incline tilted at a 12° angle above horizontal. The box is initially sliding down the incline at a speed of 2 m/s. The coefficient of kinetic friction between the box and the incline is 0.36. How far does the box slide down the incline before coming to rest? (Give answer to the nearest 0.01 meters).

C)An object weighing 4.2 N falls from rest subject to a frictional drag force given by Fdrag = bv2, where v is the speed of the object and b = 3.7 N s2/m2. What terminal speed will this object approach? (Give answer to the nearest 0.01 m/s).

Explanation / Answer

Given

A)   mass of ball m = 1.1 kg, theta = 26 degrees , velocity of the wind blowing is v= 10 m/s

here the drag force by the wind is F = ma the acceleration is a = g tan theta

           F = m*g tan theta

           F = 1.1*9.8tan26 N

           F = 5.26 N

B)box is sliding down the incline of 12 degrees above the horizontal

   initial velocity is u =2 m/s
coefficient of kinetic friction is mue =0.36

writing the force acting on the box along horizontal is

   mg sin theta - mue_k*mg cos theta = ma

   a = g( sin theta -mue_k*cos theta)

   a = 9.8(sin12 - 0.36 cos12) m/s2
   a = -1.413370 m/s2
-ve sign indicates the accelerataion is decreasing means the box is slowing down and finally stops after soem displacement

   using equations of motion v^2 -u^2 = 2as

       s = -u^2/2a
  
       s = -2^2/(2*-1.413370) m

       s = 1.4151 m

the box slide down the incline before coming to rest is 1.4151 m

C)

weight of the object w = 4.2 N

drag force Fd = bv^2

when the object reaches the terminal speed the weight = drag force

   W = Fd

   W = bv^2

   v^2 = W/b
   v = sqrt(W/b)
  
   v = sqrt(4.2/3.7) m/s
  
   v = 1.06543 m/s

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