Question

A dockworker applies a constant horizontal force of 70.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 15.0 m in 6.00 s. What is the mass of the block of ice? If the worker stops pushing at the end of 6.00 s, how far does the block move in the next 6.00 s?

Answer 1

Given F = ma = 70 N

displacement s = 15 m

time t = 6 s

initial speed u = 0m/s

from s = ut + 0.5*a*t^2

15 = 0.5 * a * 36

a = 0.83 m/s^2

a) there fore m = F/a = 70/0.83

mass m = 84.33 kg

b) After the removel of applied force the block
moves with constant velocity

after 6 s the velocity is

v = u + at

v = 0 + 0.83 * 6

v = 4.98 m/s

distance d = v* t = 4.98 * 6

v = 29.88 m

Answer 2

SOLUTION ;

Ice block moves 15 m in 6 sec. Initial speed is zero.

Hence, using formula :

s = ut + 1/2 a t^2 

=> 15 = 0 + 1/2 a (6)^2

=> 15 = 18 a 

=> a = 15/18 = 5/6 m/sec^2 

a.

So,

Mass of the block, m 

= F/a 

= 70 / (5/6) kg 

= 84 kg 

Mass of ice block is = 84 kg (ANSWER).

b.

Speed of block at t = 6 sec :

= v6 + at = 0 + 5/6 * 6 = 5 m/sec.

So distance moved in next 6 sec ;

= v6 * (t) + 1/2 (a) (t)^2

= 5 (6) + 1/2 (5/6) (6)^2

= 30 + 15

= 45 m 

Distance moved in next 6 sec = 45 m (ANSWER)

Answer 3

I am correcting part (b) as under :

b.

Speed of block at t = 6 sec :

v6 = u + at = 0 + (5/6) * 6 = 5 m/s

So distance moved in next 6 sec when force is no more applied : 

= v6  * (t) 

= 5 * 6

= 30 m 

Distance moved in next 6 sec = 30 m (ANSWER)