A 1.3-kg object is moving in the x direction at 15 m/s. Just 3.35 s later, it is moving at 28.2 m/s at 32.4° to the x axis. What are the magnitude and direction of the force applied during this time?
2007-09-09 13:31:04 · 2 answers · asked by Nathan R 1 in Science & Mathematics ➔ Physics
You can't tell, since you DIDN'T assert that the force was CONSTANT!
(All you could determine would be the magnitude of the total time-integrated force and/or impulse(s) given to the object.)
However, the x-component of the velocity change is
28.2 cos 32.4 - 15 m/s = 8.8100... m/s, its AVERAGE rate of increase in that time 2.6298... m/s^2.
The y-component is 28.2 sin 32.4 m/s = 15.1103... m/s, its average rate of increase 4.5105... m/s^2.
So if the force applied was CONSTANT during that time, its magnitude would have been given by the above averages, that is by the square root of the sum of the squares of those two accelerations times 1.3 kg, that is:
(2.6298...^2 + 4.5105...^2)^(1/2) 1.3 N
= 6.7876... N, at an agle of arctan (4.5105.../2.6298...) with the x-axis, that is at an angle of
+ 59.756 degrees relative to the + x-direction.
Live long and prosper.
2007-09-09 13:44:18 · answer #1 · answered by Dr Spock 6 · 0⤊ 0⤋
Force = Mass x Acceleration. You have the mass, that's pretty easy. Acceleration is meters per second, per second. So how many take the change in velocity and divide it by the number of seconds to get a per second change. From there it's pretty simple. As for the direction of the force, just ask yourself, which way would you push something to make it go faster, and which way to make it go slower.
2007-09-09 13:43:26 · answer #2 · answered by rohak1212 7 · 0⤊ 0⤋