An object of mass 3 kg is moving with a velocity of 5m/s along a straight. if a force of 12N is a applied for 3 seconds on the object in a perpendicular to it's direction of motion. the magnitude of velocity of the particle at the end of 3 seconds is m/s?
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We're asked to find the magnitude of the object's velocity after a force is applied for #3# #"s"# parallel to its motion.
We'll call the direction it's moving the positive #x#-axis, and the direction of the applied force the positive #y#-axis.
The components of the initial velocity are
-
#v_(0x) = 5# #"m/s"#
-
#v_(0y) = 0#
(It's moving at #5# meters per second in the straight line we called the #x#-axis.)
We know the object's mass is #3# #"kg"#, and the force applied is #12# #"N"# in the positive #y#-direction. The magnitude of the constant acceleration is thus
#a_y = (sumF_y)/m = (12color(white)(l)"N")/(3color(white)(l)"kg") = 4# #"m/s"^2#
Since this acceleration is directed upward, and the initial #y#-velocity is #0#, we can use the kinematics equation
#v_y = v_(0y) + a_yt#
to find the #y#-velocity after #3# seconds.
Plugging in known values, we have
#v_y = 0 + (4color(white)(l)"m/s"^2)(3color(white)(l)"s") = color(red)(12# #color(red)("m/s"#
No acceleration was applied in the #x#-direction, so it's #x#-velocity remains #5# #"m/s"#. The magnitude of the velocity is thus
#v = sqrt((v_x)^2 + (v_y)^2) = sqrt((5color(white)(l)"m/s")^2 + (color(red)(12)color(white)(l)color(red)("m/s"))^2)#
#= color(blue)(13# #color(blue)("m/s"#
