Question

A dock worker applies a constant horizontal force of 81.0 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 13.0 m in a time of 5.20 s .
Required:
a. What is the mass of the block of ice?
b. If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s?

Answer 1

Answer:

The correct answer is:

(a) 84.240 kg

(b) 24.038 m

Explanation:

The given values are:

Force,

F = 81.0 N

Distance,

S = 13.0 m

Time,

t = 5.20 s

As we know,

The acceleration of mass will be:

⇒  $a=\frac{2S}{t^2}$

On substituting the given values, we get

⇒     $=\frac{2\times 13.0}{(5.20)^2}$

⇒     $=\frac{26}{27.04}$

⇒     $=0.961538 \ m/s^2$

(a)

The mass of the block will be:

⇒  $m=\frac{F}{a}$

On substituting the given values, we get

⇒       $=\frac{81.0}{0.961538}$

⇒       $=84.240 \ kg$

(b)

The final velocity after a given time i.e.,

t = 5.00 s

⇒  $v=at$

On substituting the values, we get

⇒     $=0.961538\times 5.00$

⇒     $=4.8076 \ m/s$

In time, t = 5.00 s

The distance moved by the block will be:

⇒  $d=vt$

On putting the values, we get

⇒     $=4.8076\times 5.00$

⇒     $=24.038 \ m$