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?

1 Answer
Jul 4, 2017

v = 13 "m/s"

Explanation:

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"