A body of mass 3 kg is under a constant force which causes a displacement s in metres in it,given by the relation $s = \frac{1}{2} t^{2}$ , where t is in seconds. Work done by the force in 2 seconds is:-

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Answer 1

By using work energy theorem

W = $Δ$ KE = $\frac{1}{2} m v_{2}^{2}$ - $\frac{1}{2} m v_{1}^{2}$

Now s = $\frac{1}{3} t^{2} \Rightarrow$ v = $\frac{2}{3} t$

$\Rightarrow$ v $_{1}$ = 0, v $_{2}$ = $\frac{2}{3} \times$ 2 = $\frac{4}{3}$ ms $^{- 1}$

Therefore W = $\frac{1}{2} \times$ 3 $(\frac{4}{3})^{2}$ - $\frac{1}{2} \times$ 3 $\times (0)^{2}$ = $\frac{8}{3}$

Answer 2

By using work energy theorem

W =

Δ

KE =

\frac{1}{2} m v_{2}^{2}

-

\frac{1}{2} m v_{1}^{2}

Now s =

\frac{1}{3} t^{2} \Rightarrow

v =

\frac{2}{3} t

\Rightarrow

v

_{1}

= 0, v

_{2}

=

\frac{2}{3} \times

2 =

\frac{4}{3}

ms

^{- 1}

Therefore W =

\frac{1}{2} \times

3

(\frac{4}{3})^{2}

-

\frac{1}{2} \times

3

\times (0)^{2}

=

\frac{8}{3}

A body of mass 3 kg is under a constant force which causes a displacement s in metres in it,given by the relation $s = \frac{1}{2} t^{2}$ , where t is in seconds. Work done by the force in 2 seconds is:-

$\frac{5}{1 9}$ J

$\frac{3}{8}$ J

$\frac{8}{3}$ J

$\frac{1 9}{5}$ J

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?

A block of mass $m$ is pulled slowly by a minimum constant force $(F)$ on a horizontal surface through a distance $x$ . The coefficient of kinetic friction is $μ$ . Find the work done by the force $(F)$

A body of mass $2 k g$ is projected with an initial velocity of $5 m s^{- 1}$ along a rough horizontal table.The magnitude of work done on the body by the frictional forces before it is brought to rest is:

A body of mass $2 k g$ initially at rest moves under the action of an applied horizontal force of $7 N$ on a table with coefficient of kinetic friction $= 0.1$ . Compute the Work done by the applied force in $10 s$

A body of mass $40$ kg is dragged $8$ meter on a rough horizontal surface in $10$ with a constant velocity by a constant force of $20$ N, then calculate the work done by the force and total work done by all the forces.

Revise with Concepts

Work Done by a Variable Force

Work Done by a Variable Force - Problem L1

A body of mass 3 kg is under a constant force which causes a displacement s in metres in it,given by the relation s=21​t2 , where t is in seconds. Work done by the force in 2 seconds is:-

195​ J

83​ J

38​ J

519​ J

By using work energy theorem W = ΔKE = 21​mv22​ - 21​mv12​ Now s = 31​t2⇒ v = 32​t ⇒ v1​ = 0, v2​ = 32​× 2 = 34​ ms−1 Therefore W = 21​× 3(34​)2 - 21​× 3 ×(0)2 = 38​

A force acts on a 30g 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 metre and t is in second. The work done during the first 4 second is?

A block of mass m is pulled slowly by a minimum constant force (F) on a horizontal surface through a distance x. The coefficient of kinetic friction is μ. Find the work done by the force (F)

A body of mass 2kg is projected with an initial velocity of 5 ms−1 along a rough horizontal table.The magnitude of work done on the body by the frictional forces before it is brought to rest is:

A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction =0.1. Compute the Work done by the applied force in 10 s

A body of mass 40 kg is dragged 8 meter on a rough horizontal surface in 10 with a constant velocity by a constant force of 20 N, then calculate the work done by the force and total work done by all the forces.

Revise with Concepts

Work Done by a Variable Force

Work Done by a Variable Force - Problem L1

A body of mass 3 kg is under a constant force which causes a displacement s in metres in it,given by the relation $s = \frac{1}{2} t^{2}$ , where t is in seconds. Work done by the force in 2 seconds is:-

$\frac{5}{1 9}$ J

$\frac{3}{8}$ J

$\frac{8}{3}$ J

$\frac{1 9}{5}$ J

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?

A block of mass $m$ is pulled slowly by a minimum constant force $(F)$ on a horizontal surface through a distance $x$ . The coefficient of kinetic friction is $μ$ . Find the work done by the force $(F)$

A body of mass $2 k g$ is projected with an initial velocity of $5 m s^{- 1}$ along a rough horizontal table.The magnitude of work done on the body by the frictional forces before it is brought to rest is:

A body of mass $2 k g$ initially at rest moves under the action of an applied horizontal force of $7 N$ on a table with coefficient of kinetic friction $= 0.1$ . Compute the Work done by the applied force in $10 s$

A body of mass $40$ kg is dragged $8$ meter on a rough horizontal surface in $10$ with a constant velocity by a constant force of $20$ N, then calculate the work done by the force and total work done by all the forces.