Answer:
4 seconds
Explanation:
We have the following data for this exercise :
A 20.0 kg mass ⇒ [tex]m=20.0kg[/tex]
A velocity of + [tex]3.0\frac{m}{s}[/tex] ⇒ The module of this vector is the speed ⇒ We have and initial speed of [tex]3.0\frac{m}{s}[/tex]
And a constant force F with a value of 15.0 N ⇒ [tex]F=15.0N[/tex]
Let's start finding the acceleration that this force applies over the mass.
We can write the following equation :
[tex]F=m.a[/tex]
Where a is the acceleration over the mass ''m'' due to the force F.
Using this equation we can find the acceleration
[tex]15.0N=(20.0kg).a[/tex]
[tex]a=\frac{15.0N}{20.0kg}[/tex]
The unit N is equivalent to [tex]N=kg.\frac{m}{s^{2}}[/tex]
⇒
[tex]a=0.75\frac{m}{s^{2}}[/tex]
Now in order to find the time, we are going to use the following cinematic equation :
[tex]V=V0+a.t[/tex]
Where V is the speed, V0 is the initial speed and t is the time
We want the mass to stop ⇒ [tex]V=0\frac{m}{s}[/tex]
We also know the initial speed [tex]V0=3.0\frac{m}{s}[/tex]
[tex]V=V0+a.t[/tex]
[tex]0=3.0\frac{m}{s}-(0.75\frac{m}{s^{2}}).t[/tex] (I)
[tex]t=\frac{3\frac{m}{s}}{0.75\frac{m}{s^{2}}}[/tex]
[tex]t=4s[/tex]
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.
