Question

Remember to include your data, equation, and work when solving this problem.

A 20.0 kg mass moving at a velocity of + 3.0 m/s is stopped by a constant force of 15.0 N. How many seconds must the force act on the mass to stop it?

Answer 1

answer

1.5 n good look

Answer 2

4 seconds

Step-by-step explanation:

We have the following data for this exercise :

A 20.0 kg mass ⇒
`m=20.0kg`

A velocity of +
`3.0(m)/(s)` ⇒ The module of this vector is the speed ⇒ We have and initial speed of
`3.0(m)/(s)`

And a constant force F with a value of 15.0 N ⇒
`F=15.0N`

Let's start finding the acceleration that this force applies over the mass.

We can write the following equation :

`F=m.a`

Where a is the acceleration over the mass ''m'' due to the force F.

Using this equation we can find the acceleration

`15.0N=(20.0kg).a`

`a=(15.0N)/(20.0kg)`

The unit N is equivalent to
`N=kg.(m)/(s^(2))`

⇒

`a=0.75(m)/(s^(2))`

Now in order to find the time, we are going to use the following cinematic equation :

`V=V0+a.t`

Where V is the speed, V0 is the initial speed and t is the time

We want the mass to stop ⇒
`V=0(m)/(s)`

We also know the initial speed
`V0=3.0(m)/(s)`

`V=V0+a.t`

`0=3.0(m)/(s)-(0.75(m)/(s^(2))).t` (I)

`t=(3(m)/(s))/(0.75(m)/(s^(2)))`

`t=4s`

The force must act 4 seconds to stop the mass.

We add a ''-'' in equation (I) because the acceleration is opposite to the movement because it is stopping the mass.