# Finding distance using force, acceleration, time

by xmflea
Tags: acceleration, distance, force, time
 P: 44 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. Relevant 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.
HW Helper
P: 5,346
 Quote by 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. Relevant 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.
 P: 44 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
P: 44

## Finding distance using force, acceleration, time

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.

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