Question

A dockworker applies a constant horizontal force of 72.0 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 12.5 m in a time of 4.50 s . (a)What is the mass of the block of ice? (b)If the worker stops pushing at the end of 4.50 s, how far does the block move in the next 4.20s ?

Answer 1

58.54 kg

23.247 m

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

m = Mass

a) Equation of motion

`s=ut+(1)/(2)at^2\\\Rightarrow 12.5=0* 4.5+(1)/(2)* a* 4.5^2\\\Rightarrow a=(12.5* 2)/(4.5^2)\\\Rightarrow a=1.23\ m/s^2`

Force

F = ma

`m=(F)/(a)\\\Rightarrow m=(72)/(1.23)\\\Rightarrow m=58.54\ kg`

Mass of block is 58.54 kg

b)

`v=u+at\\\Rightarrow v=0+1.23* 4.5\\\Rightarrow v=5.535\ m/s`

If this velocity is constant then

Distance = Speed × Time

`\text{Distance}=5.535* 4.2\\\Rightarrow \text{Distance}=23.247\ m`

Distance the block move in the next 4.20 s is 23.247 m