Question

A dock worker applies a constant horizontal force of 80.5 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 4.50 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 ? -

Answer 1

Answer:

(a). The mass of the block of ice is 62.8 kg.

(b). The distance is 24.192 m.

Explanation:

Given that,

Horizontal force = 80.5 N

Distance = 13.0 m

Time = 4.50 s

(a). We need to calculate the acceleration of the block of ice

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$13=0+\dfrac{1}{2}\times a\times(4.50)^2$

$a=\dfrac{13\times2}{(4.50)^2}$

$a=1.28\ m/s^2$

We need to calculate the mass of the block of ice

Using formula of force

$F = ma$

$m=\dfrac{F}{a}$

Put the value into the formula

$m=\dfrac{80.5}{1.28}$

$m=62.8\ kg$

(b). If the worker stops pushing at the end of 4.50 s,

We need to calculate the velocity

Using equation of motion

$v =u+at$

Put the value into the formula

$v=0+1.28\times4.50$

$v=5.76\ m/s$

We need to calculate the distance

Using formula of distance

$v = \dfrac{d}{t}$

$d=v\times t$

Put the value into the formula

$d=5.76\times4.20$

$d=24.192\ m$

Hence, (a). The mass of the block of ice is 62.8 kg.

(b). The distance is 24.192 m.