Question

A dockworker applies a constant horizontal force of 72.0 N to a block of ice on a smooth horizontal oor. The frictional force is negligible. The block starts from rest and moves a distance 12.5 m in a time of 5.50 s.(a) What is the mass of the block of ice? (b) If the worker stops pushing after 5.50 s, how far does the block move in the next 5.70 s? [87.1 kg, 25.9 m]

Answer 1

(a).The mass of the block of ice 87.16 kg.

(b). The distance is 24.98 m.

Step-by-step explanation:

Given that,

Horizontal force = 72.0 N

Distance = 12.5 m

Time = 5.50 s

We need to calculate the acceleration

Using equation of motion

`s=ut+(1)/(2)at^2`

Where, u = initial velocity

a = acceleration

t = time

s = distance

Put the value into the formula

`12.5=0+(1)/(2)* a*(5.50)^2`

`a=(2*12.5)/((5.50)^2)`

`a=0.826\ m/s^2`

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

Using formula of force

`F=ma`

`m=(F)/(a)`

Put the value into the formula

`m=(72.0)/(0.826)`

`m=87.16\ kg`

(b). If the worker stops pushing after 5.50 s.

We need to calculate the final velocity

Using equation of motion

`v=u+at`

Put the value into the formula

`v=0+0.826*5.50`

`v=4.543\ m/s`

We need to calculate the distance

Using formula of velocity

`v=(d)/(t)`

`d=v* t`

Put the value into the formula

`d=4.543*5.50`

`d=24.98\ m`

The distance is 24.98 m.

Hence, (a).The mass of the block of ice 87.16 kg.

(b). The distance is 24.98 m.