A dockworker applies a constant horizontal force of 79.0 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 10.5 m in a time of 4.80 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 ?

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Khoso123 Khoso123 · Answer 1 · 2020-02-11T05:17:36+01:00

Answer:

(a) mass of block ice=86.7 kg

(b) Distance=17.2179 m

Explanation:

Given data

For part (a)

First find acceleration then mass by ΣFx=ma

Let +x be the direction of force

ΣFx=79 N

x-x₀=10.5 m

t=4.80 s

Initial velocity V₀=0 m/s

[tex]x-x_{o}=v_{o}t+(1/2)at^{2}\\ 10.5m=0+(0.5)a(4.80s)^{2}\\ 10.5m=11.52a\\a=10.5/11.52\\a=0.911m/s^{2}[/tex]

Now ΣFx=ma

So

[tex]79N=m(0.911m/s^{2} )\\m=79/0.911\\m=86.7kg[/tex]

For part (b)

First calculate the speed at the end period 4.50s of applied force

Then use the ending velocity as initial velocity in the second part of motion

After first 4.50 seconds

[tex]v=v_{o}+at\\ v=0+(0.911m/s^{2} )(4.50s)\\v=4.0995m/s\\[/tex]

acceleration ax=0 m/s²

velocity v=constant

So

[tex]x-x_{o}=vt+(1/2)a_{x}t^{2}\\x-x_{o}=(4.0995m/s)(4.20s)+0\\x-x_{o}=17.2179m[/tex]