Question

A time dependent force, F = (8.00i – 4.00t j)N
(where t is in seconds), is applied to a 2.00 kg object
initially at rest.
(a) At what time will the object be moving with a
speed of 15.0 m/s?
(b) How far is the object from its initial position when
its speed is 15.0 m/s?​

Answer 1

Given that,

Force $= (8.00i-4.00tj)\ N$

Mass = 2.00 kg

Speed = 15.0 m/s

(a). We need to calculate the time

Using formula of force

$F=ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{8.00i+(4.00t)j}{2}$

$a=4.00i+(2.00t)j$

On integration

$\dfrac{dv}{dt}=4.00i+(2.00t)$

$v=(4.00t)i+(t^2)j$

Put the value of v

$15=(4.00t)i+(t^2)j$

The magnitude of velocity

$15^2=(4.00t)^2+(t^2)^2$

$t^4+16t^2=225$

$t^2+16t=225$

$t=9\ sec$

(b). We need to calculate the distance the object from its initial position

Using formula of distance

$\dfrac{dx}{dt}=(4.00t)i+(t^2)j$

On integration

$x=(2.00t^2)i+\dfrac{t^3}{3}j$

Put the value into the formula

$x=(2.00\times9^2)i+(\dfrac{(9)^2}{3}j)$

$x=\sqrt{(2.00\times9^2)^2+(\dfrac{(9)^2}{3})^2}$

$x= 164.23\ m$

Hence, (a). The time is 9 sec.

(b). The distance the object from its initial position is 164.23 m.