Question

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?

Answer 1

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 is the applied force
• m is the mass of the puck
• a is the acceleration of the puck

F = mv/t

mv = Ft

v = Ft / m

where;

• v is the speed of the puck at the time

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

Answer 2

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.