Question

The following time dependent force is applied to a 2 kg block that is initially at rest on a frictionless and horizontal surface. F open parentheses t close parentheses equals open parentheses 4 straight N over straight s close parentheses t What is the displacement (in m) of the block after 3 s?

Answer 1

Answer:

The displacement of the block is 9 m.

Explanation:

Given that,

Mass of block = 2 kg

Time = 3 sec

The force is

$F(t)=4t$

The acceleration is

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

Put the value into the formula

$a=\dfrac{4t}{2}$

$a=2t$

We need to calculate the velocity

Using formula of acceleration

$v=\int_{0}^{t}{a dt}$

Put the value into the formula

$v=\int_{0}^{3}(2t dt)$

$v=(\dfrac{2t^2}{2})_{0}^{3}$

$v=(t^2)_{0}^{3}$

$v=9\ m/s$

We need to calculate the displacement

Using formula of velocity

$s=\int_{0}^{3}(v dt)$

$s=(t^2)_{0}^{3}$

$s=(\dfrac{t^3}{3})_{0}^{3}$

$s=9\ m/s$

Hence, The displacement of the block is 9 m.

Answer 2

Answer:

$s=9\ m$

Explanation:

Given:

mass of the block,$m=2\ kg$

force as a function of time, $F(t)=4t$

time of observation, $t=3\ s$

initial velocity of the block, $u=0\ m.s^{-1}$

Now the acceleration:

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

$a=\frac{4t}{2}$

$a=2t$

Now the related velocity:

$v=\int {a} \, dt$

$v=\int 2t\ dt$

$v=t^2$

Integrating again we get the displacement:

$s=\int\limits^3_0 v\ dt$

$s=\int\limits^3_0 t^2\ dt$

$s=\frac{t^3}{3}\vert^3_0$

$s=9\ m$