Question

A crate with mass 27.0kg initially at rest on a warehouse floor is acted on by a net horizontal force of 126N .

Part A What acceleration is produced? a =   m/s2   SubmitMy AnswersGive Up Part B How far does the crate travel in 10.5s ? x = m SubmitMy AnswersGive Up Part C What is its speed at the end of 10.5s ? v = m/s

Answer 1

apply the relation between force and accleration as F= ma

where m is mass

a is accleration and

F is force

so as we need

part A: accleration a = F/m

a = 126/27

a = 4.667 m/s^2 ------------<<<<<<<<<<<<<<<<Answer to part A

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part B :

use the kinematic equation S = ut + 0.5 at^2

where a is accleration   = 4.67 m/s^2

t is time = 10.5 secs

as it starts from rest initial velocity u =   0

so

S = x =   distance travelled = 0.5 * 4.67 * 10.5 *10.5

x   = 257.44 meters------------<<<<<<<<<<<<<<<<Answer to part B

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part C:

use again the Kinematic equation V = u + at

V = 0 + (4.67* 10.5)

V = 49.035 m/s----------------------<<<<<<<<<<<<<<<<<Answer to part C