A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.25 s?

Question

asked Mar 21, 2022 191k views

2 Answers

Tommy Strand · Answer 1 · 2022-03-22T15:43:48+0000

The distance traveled by the puck in 2.25 seconds is 8.4 m. (Option C).

How to calculate the distance moved by the puck?

The distance traveled by the puck in 2.25 seconds is calculated by applying the following formula as shown below;

F = ma

where;

F = mv/t

mv = Ft

v = Ft / m

where;

The speed is calculated as follows;

v = ( 5 N x 1.5 s ) / ( 2 kg )

v = 3.75 m/s

The distance traveled by the puck at 2.25 s is calculated as

D = 3.75 m/s x 2.25 s

D = 8.4 m

Greatertomi · Answer 2 · 2022-03-27T02:39:04+0000

Answer:

d = 6.32 m

Step-by-step explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

$a=(F)/(m)\\\\a=(5)/(2)\\\\a=2.5\ m/s^2$

Let d is the distance moved in 2.25 s. Using second equation of motion,

$d=ut+(1)/(2)at^2\\\\d=0+(1)/(2)* 2.5* (2.25)^2\\\\d=6.32\ m$

So, it will move 6.32 m from rest in 2.25 seconds.