Question

A dockworker applies a constant horizontal force of 82.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 13.0 m in a time of 4.90 s(a)What is the mass of the block of ice?----- I found this to be 56.9kg and got it right.(b)If the worker stops pushing at the end of 4.50 s, how far does the block move in the next 4.20s ? ------ I tried using [distance=0.5at^2] but it says its wrong. how do you do this question?

Answer 1

Answer:

75.7238461542 kg

20.4664723031 m

Explanation:

F = Force = 82 N

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 13=0\times t+\frac{1}{2}\times a\times 4.9^2\\\Rightarrow a=\frac{13\times 2}{4.9^2}\\\Rightarrow a=1.08288213244\ m/s^2$

Force is given by

$F=ma\\\Rightarrow m=\dfrac{F}{a}\\\Rightarrow m=\dfrac{82}{1.08288213244}\\\Rightarrow m=75.7238461542\ kg$

The mass of the block is 75.7238461542 kg

$v=u+at\\\Rightarrow v=0+1.08288213244\times 4.5\\\Rightarrow v=4.87296959598\ m/s$

Distance moved in the next 4.2 seconds

$d=vt\\\Rightarrow d=4.87296959598\times 4.2\\\Rightarrow d=20.4664723031\ m$

The distance moved is 20.4664723031 m