F = ma
0.25N = 0.160kg * a
solve for a
a = 0.25(kg m/s2)/0.160 kg = 1,5625 m/s2
If the puck is at rest and then accelerates at a constant rate the position is:
x = (1/2)at2 = (1/2)1,5625 m/s2 * (2 s)2
x = 3.125 meters away from the start position
Hope this helps
the velocity is given by
v = at
so since a = 1,5625 m/s2
v = at = 1,5625 m/s2 * 2s = 3.125 m/s
WARNING the fact that these two problems have the same numerical value is only a cooincidence!
I guess there is the bid charge (at least on your end) for every question you ask.
Sorry I don't make the rules.
I guess that means you have to be careful what you ask.
I have seen people post several problems as part of a single question. The risk of doing that is if the assignment is too long you may not get a response.
Still $15 for that 2 line answer IS a bit steep.
Next time I'd post both questions together as a single question.
Again my apologies
throw rock up into air
v = at + vo
where a is the acceleration and vo is the initial velocity
I like to look at this like this. The rock spends 1/2 it's time going up and 1/2 it's time returning to the same level
so the time going up is 1.51s/2 = 0.755 s
The rock decellerates from 13.0 m/s to 0 m/s in this amount of time so
a = dv/dt = (13.0m/s)/(0.755 s2) = 17.22 m/s2
(about 1.75 x earths gravity)
What does the rock weigh ?
in Newtons it would be F = mgplanet x = 5.24kg*17.22 m/s2
= 90.2 N
hope this helps