Question

A dockwater applies a constant horizontal force of 80.0 N to a block of ice ona smooth horizontal floor. The frictional force is negligible. The block starts from rest and moces 11.0 m in 5.00s. (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 5.00s?

Answer 1

Answer 2

SOLUTION ;

Ice block moves 11.0 m in 5 sec. Initial speed is zero.

Hence, using formula :

s = ut + 1/2 a t^2 

=> 11.0 = 0 + 1/2 a (5)^2

=> 11.0 = 12.5 a 

=> a = 11.0 / 12.5 = 0.88 m/sec^2 

a.

So,

Mass of the block, m 

= F/a 

= 80 / 0.88  kg 

= 90.91 kg 

Mass of ice block is = 90.91 kg (ANSWER).

b.

Speed of block at t = 5 sec :

v5 = u + at = 0 + 0.88 * 5 = 4.4 m/s

So distance moved in next 5 sec when force is no more applied : 

= v5  * (t) 

= 4.4 * 5

= 22 m 

Distance moved in next 5 sec = 22 m (ANSWER)