Question

A stone of 10g thrown with a velocity of 10m/s across a floor travels through a distance before coming to rest in 0.02s. Calculate the (i) distance travelled and (ii) the force exerted by the floor on the stone.

Answer 1

Answer:

let the mass 10 g=0.01kg

initial velocity (u)= 10m per s

finial velocity (v) = 0m per s

time taken 0.02s

now accelartion is v-u/t

om per s -10m/s/o.o2s = -500m/s^2

distance = 2as=v^2-u^2

2*-500m/s^2 *s=0m^2/s^2-100m^2/s^2

0.1m distance

now force F =MA

F = o. o1 kg* -500m/s^2= -5kgms^2 or 5 N

so the Stone exerted force on floor is 5N and due to action and reactions the same force exerts floor on stone which is -5N in opposite directions the force exerted by floor is kinetic friction