A dockworker applies a constant horizontal force of 60.0 N to a block of ice on a smooth horizontal floor.?

Question

The frictional force is negligible. The block starts from rest and moves 14.0 m in 9.00 s.
(a) What is the mass of the block of ice?
kg

(b) If the worker stops pushing after 9.00 s, how far does the block move in the next 9.00 s?
m

Answer 1

using F=m*a and
v(t)=a*t and
y(t)=.5*a*t^2

60=m*a
a=60/m
14=.5*a*81
plugging in a
m=30*81/14

for part b
the speed after the first 9 seconds is
v(9)=9*60/m
and for the next 9 seconds this is v0
so
d=81*60/m
=14*81*60/(30*81)
=28 m

j

Answer 2

supposing that F= 76N, S = 10m and t = 5.5sec , it fllows that S = u t + a million/2 a t^2 , substituting the values and simplifying we get, a = 0.sixty six m/s^2 then mass m = F /a = seventy six / 0.sixty six = one hundred fifteen.15 kg on the top of five.5 sec the speed obtained via block is V = 0 + at = 0.sixty six x 5.5 = 3.sixty 3 m/sec consequently in next 5.6 sec, the area travelled via it somewhat is going to be S* = V x t = 3.sixty 3 x 5.6 = 20.326 m