Newton's second law and direction of force

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 3K views
neutron star
Messages
78
Reaction score
1

Homework Statement


A particle is traveling in a straight line at a constant speed of 22.1 m/s. Suddenly, a constant force of 12.6 N acts on it, bringing it to a stop in a distance of 55.3 m.


Homework Equations





The Attempt at a Solution



(a) What is the direction of the force?
perpendicular to the direction of the particle's motion
*opposite the direction of the particle's motion
the same as the direction of the particle's motion
none of the above

(b) Determine the time it takes for the particle to come to a stop.
___ s

(c) What is its mass?
___ kg

F=ma
[tex]V_f=0[/tex]
[tex]V_f ^2 = V_i ^2 + 2a(X-X_i)[/tex]
[tex]0^2=(22.1m/s)^2 + 2a(55.3m)[/tex]
[tex]0=(22.1m/s)+at[/tex]
 
Physics news on Phys.org
neutron star said:
F=ma
[tex]V_f=0[/tex]
[tex]V_f ^2 = V_i ^2 + 2a(X-X_i)[/tex]
[tex]0^2=(22.1m/s)^2 + 2a(55.3m)[/tex]
[tex]0=(22.1m/s)+at[/tex]


Yes so the deceleration is ? And the time taken is?
 
[tex]0=22.1m/s^2+2a(55.3)[/tex]
[tex]-110.6a=488.41m/s[/tex]
[tex]a=-4.416m/s[/tex]

[tex]0=22.1m/s^2+(-4.416m/s)t[/tex]
[tex]22.1m/s^2/4.416m/s=5.004s[/tex]
[tex]t=5.004s[/tex]

Is this right?

How do I get mass from this? F=ma right, so m=F/a or 12.6=m(-4.416)
or m=12.6/-4.416. But it doesn't seem to work...
 
Last edited: