Question

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?

Answer 1

`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"`