Question

Help PLEASE!
A particle is traveling in a straight line at a constant speed of 20.1 m/s. Suddenly, a constant force of 11.3 N acts on it, bringing it to a stop in a distance of 68.9 m.
1.) Determine the time it takes for the particle to come to a stop.
=>__________s.
2.) What is its mass?
=>__________kg.

Answer 1

1) apply \[v^2 = v_0^2 + 2as\] at the time the particle stop, \(v =0\) there fore \(v_0^2 = -2as\rightarrow a = -\dfrac{v_0^2}{2s}=-\dfrac {20.1^2}{2* 68.9}= -2.93 (m/s^2)\)
the minus sign shows that it is decelerate
you have a, you have \(v_0\) you can calculate time by formula \(v = v_0 + at\)
where v =0 , \(t =-\dfrac{v_0}{a}= \dfrac{-20.1}{-2.93}= 6.9 (s)\)
2) for mass, pay attention into the given condition, you have formula F = ma \(\rightarrow m =\dfrac{F}{a}\)
\(m = \dfrac{11.3}{2.93}= 3.85kg

Answer 2

@ybarrap check my stuff, please. I am new in physics field, :)

Answer 3

Wow, that was so much easier to understand. Thank you so much!!!! (: @Loser66

Answer 4

glad to hear

Answer 5

Perfect!