Question

A dockworker applies a constant horizontal force of 73.0 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 4.50 s.(a)What is the mass of the block of ice?----- I found this to be 56.9kg and got it right.(b)If the worker stops pushing at the end of 4.50 s, how far does the block move in the next 4.20s ?

Answer 1

(a) Mass of the car will be 56.8285 kg

(b) Distance traveled in next 4.20 sec will be 33.9619 m

Step-by-step explanation:

We have given horizontal force F = 73 N

As it is given that the block starts from the rest so initial velocity u = 0 m/sec

Distance traveled s = 13 m

Time t = 4.5 sec

(a) From second equation of motion
`s=ut+(1)/(2)at^2`

`13=0* 4.5+(1)/(2)* a* 4.5^2`

`a=1.2839m/sec^2`

Now according to newton second law force F = ma

So mass
`m=(F)/(a)=(73)/(1.2839)=56.8580kg`

(b) Velocity after 4.20 sec v = u+at

v=0+1.2839×4.20 = 5.39 m/sec

So initial velocity for second case = 5.39 m/sec

From second equation of motion
`s=ut+(1)/(2)at^2`

`s=5.39* 4.20+(1)/(2)* 1.2839* 4.20^2=22.638+11.3239=33.9619m`