Question

18. A net force of 1.0N acts on a 4.0-kg object, initially at rest, for 4.0 seconds. What is the distance the object moves during the same time? Be sure to show all work to support your answer.

Answer 1

acceleration = force/mass
1/4
s=ut+1/2at^2
0*4+1/2*1/4*4^2
1/2*1/4*16
1/2*4
=2

Answer 2

Distance moved, s = 2 meters

Step-by-step explanation:

It is given that,

Net force acting on object, F = 1 N

Mass of object, m = 4 kg

We have to find the distance moved by object during 4 seconds.

From Newton's second law :
`a=(F)/(m)`

`a=(1)/(4)=0.25\ m/s^2`

Using second equation of motion to find distance :

`s=ut+(1)/(2)at^2`

s is distance, u = 0 because initially the object is at rest

`s=(1)/(2)at^2`

`s=(1)/(2)* 0.25* (4)^2`

s = 2 m

Hence, the distance moved by particle in 4 seconds is 2 meters.