Finding distance using force, acceleration, time

[xmflea]

A dockworker applies a constant horizontal force of 89.0N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 12.5m in a time 5.10s .
A)What is the mass of the block of ice?
B)If the worker stops pushing at the end of 5.10s , how far does the block move in the next 5.30s ?

2. Homework Equations : F=ma


3. The Attempt at a Solution : well i was able to solve part A by using the equation
deltaD=(v1+v2)/deltaT to find v2. and then i used a=(v2-v1)/deltaT and plugged it into F=ma to find the mass.
For part B, i tried using the same equation for finding distance, however it could not be done cause i do not know how to find v2, and if i plug in 0 as v2, i get a wrong answer.

so basically i need help in solving part B of this problem. thanks.

 

[LowlyPion]

A dockworker applies a constant horizontal force of 89.0N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 12.5m in a time 5.10s .
A)What is the mass of the block of ice?
B)If the worker stops pushing at the end of 5.10s , how far does the block move in the next 5.30s ?

2. Homework Equations : F=ma


3. The Attempt at a Solution : well i was able to solve part A by using the equation
deltaD=(v1+v2)/deltaT to find v2. and then i used a=(v2-v1)/deltaT and plugged it into F=ma to find the mass.
For part B, i tried using the same equation for finding distance, however it could not be done cause i do not know how to find v2, and if i plug in 0 as v2, i get a wrong answer.

so basically i need help in solving part B of this problem. thanks.

Welcome to PF.

Don't you want to consider the relationship between x, a, and t?

x= 1/ 2 * a * t²

If you find a, then F = ma gives you the mass as I think you already know.

Part b can be determined by finding the V from

V² = 2*a*x

That times t gives you distance.

 

[xmflea]

you mean V^2=2 x .96 x 12.5? that gives me V=4.9, which i already had to use for part A. and if i do 4.9 x 0.2s. i get .98m, which is not the right distance

 

[xmflea]

oh wait nvm, i was reading the question wrong, i thought it was asking what the distance was after 5.3 seconds from 5.1 seconds. i didnt know it meant it was asking the distance after 5.3 MORE seconds. thanks, i got the right answer now.