Question

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

Answer 1

Answer 

1.5 n good look

Answer 2

Answer:

4 seconds

Explanation:

We have the following data for this exercise :

A 20.0 kg mass ⇒ $m=20.0kg$

A velocity of + $3.0\frac{m}{s}$ ⇒ The module of this vector is the speed ⇒ We have and initial speed of $3.0\frac{m}{s}$

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

$15.0N=(20.0kg).a$

$a=\frac{15.0N}{20.0kg}$

The unit N is equivalent to $N=kg.\frac{m}{s^{2}}$

⇒

$a=0.75\frac{m}{s^{2}}$

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 ⇒ $V=0\frac{m}{s}$

We also know the initial speed $V0=3.0\frac{m}{s}$

$V=V0+a.t$

$0=3.0\frac{m}{s}-(0.75\frac{m}{s^{2}}).t$ (I)

$t=\frac{3\frac{m}{s}}{0.75\frac{m}{s^{2}}}$

$t=4s$

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.