132

You visited us 132 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Relation between Work Done and PE

A force acts ...

Question

A force acts on a 30 gm 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 metres and t is in seconds. The work done during the first 4 seconds is

5.28 J

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

450 mJ

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

490 mJ

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

530 mJ

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

Open in App

Solution

The correct option is A
5.28 J

$v = \frac{d x}{d t} = 3 - 8 t + 3 t^{2}$

$∴ v_{0} = 3 m / s a n d v_{4} = 19 m / s$

$W = \frac{1}{2} m (v_{4}^{2} - v_{0}^{2})$ (According to work energy theorem)

$= \frac{1}{2} \times 0.03 \times (19^{2} - 3^{2}) = 5.28 J$

Suggest Corrections

Similar questions

A force acts on a 30 gm 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 metres and t is in seconds. The work done during the first 4 seconds is

Q. A force acts on a

30 g m

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 metres and

t

is in seconds. The work done on the particle during the first

4

second 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 metre and t is in second. The work done during the first

4

second 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

Join BYJU'S Learning Program

A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by x=3t−4t2+t3, where x is in metres and t is in seconds. The work done during the first 4 seconds is

The correct option is A 5.28 J v=dxdt=3−8t+3t2 ∴v0=3m/s and v4=19m/s W=12m(v24−v20) (According to work energy theorem) =12×0.03×(192−32)=5.28 J

A force acts on a 30 gm 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 metres and t is in seconds. The work done during the first 4 seconds is

The correct option is A
5.28 J

$v = \frac{d x}{d t} = 3 - 8 t + 3 t^{2}$

$∴ v_{0} = 3 m / s a n d v_{4} = 19 m / s$

$W = \frac{1}{2} m (v_{4}^{2} - v_{0}^{2})$ (According to work energy theorem)

$= \frac{1}{2} \times 0.03 \times (19^{2} - 3^{2}) = 5.28 J$