Under the action of a force- a 2 kg body moves such that its position x as a function of time t is given by x - t33- where x is in meter and t in second- The work done by the force in the first two seconds is

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

Byju's Answer

Standard XII

Physics

Constrained Motion : General Approach

Under the act...

Question

Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by $x = \frac{t^{3}}{3},$ where x is in meter and t in second. The work done by the force in the first two seconds is

1600 J

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

160 J

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

16 J

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

1.6 J

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

Open in App

Solution

The correct option is C 16 J
$v = \frac{d x}{d t} = t^{2}$
At t = 0, v = 0,
At t = 2 s, v = 4 m/s
From work energy theorem,
W = change in kinetic energy
$K_{f} - K_{i}$
$= \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$= \frac{1}{2} \times 2 \times (16 - 0)$
$= 16 J$

Suggest Corrections

Similar questions

Q. Under the action of a force, a

2 kg

body moves such that its position

x

as a function of time

t

is given by

x = \frac{t^{3}}{3}

, where

x

is in

meter

and

t

second

. The work done by the force in the first two seconds is (Give an integer value)

Q. Under the action of a force, a

1 k g

body moves, such that its position

x

as a function of time

t

is given by

x = \frac{t^{3}}{2}

where

x

is in meter and

t

is in second. The work done by the force in first

3

second is

Q. Under the action of a force, a

2 kg

body moves such that its position

x

as a function of time t is given by

x = \frac{t^{3}}{3}

, where

x

is in meter and

t

in second. The work done by the force in the first two seconds is

Q. Under the action of a force, a

2 k g

body moves such that its position

x

(in meters) as a function of time

t

(in seconds) is given by

x = t^{2} / 2

. The work done by the force in the first

5

seconds is:

Q. Under the action of a force, a

2 l g

body moves such that it position

x

as a function of time is given by

x = \frac{t^{3}}{3}

, where

x

is in metre and

t

in seconds. The work done by the force in the first two seconds is

Join BYJU'S Learning Program

Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by x=t33, where x is in meter and t in second. The work done by the force in the first two seconds is

The correct option is C 16 Jv=dxdt=t2 At t = 0, v = 0, At t = 2 s, v = 4 m/s From work energy theorem, W = change in kinetic energy Kf−Ki =12m(v2f−v2i) =12×2×(16−0) =16 J

Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by $x = \frac{t^{3}}{3},$ where x is in meter and t in second. The work done by the force in the first two seconds is

The correct option is C 16 J
$v = \frac{d x}{d t} = t^{2}$
At t = 0, v = 0,
At t = 2 s, v = 4 m/s
From work energy theorem,
W = change in kinetic energy
$K_{f} - K_{i}$
$= \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})$
$= \frac{1}{2} \times 2 \times (16 - 0)$
$= 16 J$