The displacement x of a particle at time t moving under a constant force is t- x -3- x in metre- t in second- Find the work done by the interval from t-0 to t-6 s-

Given, t=√( x ) +3 ⇒ x=( t 3 ) nbsp;^ 2 ⇒ v=2( t 3 ) nbsp; nbsp; nbsp; nbsp; ( ∵ v= dx dt nbsp;) At t=0, v= 6 ms ^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Gravtiational PE

The displacem...

Question

The displacement $x$ of a particle at time $t$ moving under a constant force is $t = \sqrt{x} + 3$ , $x$ in metre, $t$ in second. Find the work done by the interval from $t = 0$ to $t = 6 s$ .

Open in App

Solution

Given, $t = \sqrt{x} + 3$
$\Rightarrow x = {(t - 3)}^{2}$
$\Rightarrow v = 2 (t - 3)$ $(∵ v = \frac{d x}{d t})$
At $t = 0, v = - 6 {m s}^{- 1}$
At $t = 6 s, v = 6 {m s}^{- 1}$
Work done $=$ Change in kinetic energy $= 0$ .

Suggest Corrections

Similar questions

Q. The displacement

x

of a particle moving in one dimension, under the action of a constant force is related to time

t

by the equation

t = \sqrt{x} + 3

where x is in metre and t in second. Calculate the work done by the force in the first

6 s

J o u l e

Q. The displacement

x

of a particle moving in one dimension, under the action of a constant force is related to time

t

by the equation

t = \sqrt{x} + 3

where x is in metre and t in second. Calculate the velocity at any

t

The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation, t = $\sqrt{x}$ + 3 where x is in meters and t is in seconds. The work done by the force in the first 6 seconds is

Q. The displacement

x

of a particle moving in one dimension, under the action of a constant force is related to time

t

by the equation

t = \sqrt{x} + 3

where x is in metre and t in second. Calculate the displacement of the particle when its velocity is

z e r o

Q. The displacement

x

of a particle moving in one dimension under the action of a constant force is related to time

t

by the equation

t - 3 = \sqrt{x}

where

x

is in metres and

t

is in seconds. Find the displacement of the particle when its velocity is zero.

Join BYJU'S Learning Program

The displacement x of a particle at time t moving under a constant force is t=√x+3, x in metre, t in second. Find the work done by the interval from t=0 to t=6s.

Given, t=√x+3⇒x=(t−3)2⇒v=2(t−3) (∵v=dxdt)At t=0,v=−6ms−1At t=6s,v=6ms−1Work done = Change in kinetic energy =0.

The displacement $x$ of a particle at time $t$ moving under a constant force is $t = \sqrt{x} + 3$ , $x$ in metre, $t$ in second. Find the work done by the interval from $t = 0$ to $t = 6 s$ .

Given, $t = \sqrt{x} + 3$
$\Rightarrow x = {(t - 3)}^{2}$
$\Rightarrow v = 2 (t - 3)$ $(∵ v = \frac{d x}{d t})$
At $t = 0, v = - 6 {m s}^{- 1}$
At $t = 6 s, v = 6 {m s}^{- 1}$
Work done $=$ Change in kinetic energy $= 0$ .