Question

A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.00 s.

Answer 1

The net force acting on this object is 180.89 N.

Step-by-step explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of
`x= 5t^3+1`

Coordinate of position of
`y=3t^2+2`

Time = 2.00 s

We need to calculate the acceleration

`a = (d^2x)/(dt^2)`

For x coordinates

`x=5t^3+1`

On differentiate w.r.to t

`(dx)/(dt)=15t^2+0`

On differentiate again w.r.to t

`(d^2x)/(dt^2)=30t`

The acceleration in x axis at 2 sec

`a = 60i`

For y coordinates

`y=3t^2+2`

On differentiate w.r.to t

`(dy)/(dt)=6t+0`

On differentiate again w.r.to t

`(d^2y)/(dt^2)=6`

The acceleration in y axis at 2 sec

`a = 6j`

The acceleration is

`a=60i+6j`

We need to calculate the net force

`F = ma`

`F = 3.00*(60i+6j)`

`F=180i+18j`

The magnitude of the force

`|F|=√((180)^2+(18)^2)`

`|F|=180.89\ N`

Hence, The net force acting on this object is 180.89 N.