Question

A dock worker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in the first 5.00 s. What is the mass of the block of ice

Answer 1

The mass of the block of ice is 90.91 kg.

Step-by-step explanation:

The expression for the equation of motion is as follows;

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

Here, u is the initial speed, a is the acceleration, t is the time and s is the distance.

Calculate the acceleration of the given block.

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

Put t= 5 s, s= 11 m and u= 0 in the above expression.

`11=(0)t+(1)/(2)a(5)^2`

`a= 0.88 ms^(-2)`

The expression for the force in terms of mass and acceleration is as follows;

F= ma

Here, F is the force and m is the mass.

Put F= 80 N and
`a= 0.88 ms^(-2)`.

80= m(0.88)

m=90.91 kg

Therefore, the mass of the block of ice is 90.91 kg.