Question

A 50 kg stone falls into a mud pot from a height of 6 m. It sinks 0.5 m below the mud surface

before coming to rest. Assuming a constant frictional force between the mud and stone

brought the stone to rest, find the magnitude of that force.​

Answer 1

Answer:

Explanation:

We know v^2-u^2=-2as (here v=0 & `a` which is due to friction opposes the motion hence -ve & as it is freefall upto point of contact with mud so u=√2gh=√2*10*6)So a=u^2/2sGiven, s=0.5,So a=(√2*10*6)^2/2*0.5 =120We also know, F=ma=50*120=6000N

Answer 2

Here we consider 2 cases.. Case 1 when object fall till the surface ..till it accelerates.. So u=0m/s,g=9.8m/s^2 ,s=6m..... v^2 - u^2=2gs. v^2-0=2*9.8*6 v^2=117.6m^2/s^2.Case 2... Now object deaccelerate..So this time final velocity become 0 .. and initial velocity will be final velocity of case 1...So .. v^2-u^2=2as , 0-117.6=2*a*0.5 , a= 117.6 m/s^2 ...Force acting during deaccelerating is... f = m *a = 50* (-117.6) = 5880 N..