Question

5 points Consider a flat metal plate of negligible thickness and area of each face A = 0.5 m2, and mass m = 10.0 kg The plate is placed on a large flat sheet of ice, such as a frozen lake surface. Due to the weight of the plate, some of the ice under the plate melts to form a layer of water of thickness h=5.0e-4 m. At time t = Os (ie. initial time), a constant force horizontal F = 20.0N is applied on the plate so that the plate starts sliding on the ice. After 10.0s, the horizontal force F is removed and the plate is allowed to continue sliding along the ice with the viscous friction as the only external horizontal force on it To find the distance in m covered in the time starting when the force is removed till the plate comes to rest The coefficient of viscosity of water = 0.0018 pas

Answer 1

The thickness of water layer is so small that we can assume the velocity variation between  the layer to be linear

Viscous force acting on the plate = applied force = (1)

u**A

where, $\mu$ = dynamic viscosity = 0.0018 Pa-s

u = velocity of plate

h = thickness of water layer = 0.0005 m

A = area of plate = 0.5 m²

From (1), 20 = 0.0018*u*0.5/0.0005

=> u = 11.11 m/s

acceleration of plate, a = F/m

= 20/10

= 2 m/s²

Now once the external force is removed, plate moves with constant acceleration

Final velocity of plate, v = 0

Distance travelled by plate before coming to rest (s) is obtained from Newtons law of motion

  

v=u+2*a*s

=> 0 = 11.11² + 2*2*s

or s = 30.86 m