A force acts on a $1$ kg body such that the distance moved varies as a function of time according to the equation $X = 4 t^{2} + 3 t + 5$ , where $X$ is in the meters and $t$ in sec. What is the work done by the force in first three seconds?

120 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

240 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

320 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

360 J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 360 J
Given
$m = 1$ $K g$ , $t = 3 s$ and the distance moved varies as function of time according to the equation, $X = 4 t^{2} + 3 t + 5$
For $t = 3$
$X = 4 (3)^{2} + 3 (3) + 5$
$X = 50 m$
And For $t = 0$ ,
$X = 4 (0)^{2} + 3 (0) + 5$
$X = 5 m$
$∴ Δ X = 45 m$
Now, force is given by
$F = m a = m \frac{d^{2} X}{d t^{2}}$
$F = 1 \times \frac{d^{2} (4 t^{2} + 3 t + 2)}{d t^{2}}$
$F = 8$
now,
$W = F . Δ X = (8) (45) = 360 J$

Suggest Corrections

Similar questions

The distance $x$ moved by a body of mass 0.5 kg due to a force varies with time $t$ as
$x = 3 t^{2} + 4 t + 5$
where $x$ is expressed in metre and t in second. What is the work done by the force in the first 2 seconds?

Q. A force acts on a

20 g

particle in such a way that the position of the particle as a function of time is given by

x = 3 t - 4 t^{2} + t^{3}

, where x is in meters and

t

is in seconds. The work done during the first

4 s e c

is:

Q. A force acts on a

30 g

particle in such a way that the position of the particle as a function of time is given by

x = 3 t - 4 t^{2} + t^{3}

, where

x

is in meters and

t

is in seconds. The work done during the first

4

seconds is

Q. A force acts on a

3 g

particle in such a way that position of the particle as a function of time is given by

x = 3 t - 4 t^{2} + t^{3}

, where x is in metre and t is in sec. The work done during the first

4 s

Join BYJU'S Learning Program

A force acts on a 1 kg body such that the distance moved varies as a function of time according to the equation X=4t2+3t+5, where X is in the meters and t in sec. What is the work done by the force in first three seconds?

The correct option is D 360 JGivenm=1 Kg, t=3s and the distance moved varies as function of time according to the equation, X=4t2+3t+5For t=3 X=4(3)2+3(3)+5X=50mAnd For t=0, X=4(0)2+3(0)+5X=5m∴ΔX=45mNow, force is given byF=ma=md2Xdt2F=1×d2(4t2+3t+2)dt2F=8now,W=F.ΔX=(8)(45)=360J

A force acts on a $1$ kg body such that the distance moved varies as a function of time according to the equation $X = 4 t^{2} + 3 t + 5$ , where $X$ is in the meters and $t$ in sec. What is the work done by the force in first three seconds?