Question

14. A particle of mass 3 kg is moving along x-axis and its position at time t is given by equation x = (2t^2 + 5) m.
Work done by all the forces acting on it in time interval t = 0 to t = 3 s is
(1) 144 J
(2) 72J
(3) 108 J
(4) 216 J​

Answer 1

Answer:

By the given eqn x=2t^2+5 We can get the acceleration by double differentiating the eqn d^2x/dt^2 = acceleration= 4m/s^2 Now by 2nd low of equation S=ut+1/2at^2 Let initial velocity be = 0 Then we getS=1/2*4*9= 18m ( t=3)Now work done = FS =m*a*sW= 3*4*18= 216 joules

Answer 2

Answer:

Work done, W = 216 J      

Explanation:

It is given that,

Mass of the particle, m = 3 kg

Its position at time t is given by :

$x=2t^2+5$

We know that, work done is given by :

$W=F.s$

$W=ma.s$...........(1)

s is the displacement in time interval t = 0 to 3 s.

Velocity, $v=\dfrac{dx}{dt}$

$v=\dfrac{d(2t^2+5)}{dt}$

$v=4t$

at t = 3 s

$v=4\times 3=12\ m/s$

Acceleration, $a=\dfrac{v-u}{t}$

$a=\dfrac{12-0}{3}$

$a=4\ m/s^2$

Displacement in time interval t = 0 to 3 s is :

$s=ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}\times 4\times (3)^2$

s = 18 m

Equation (1) becomes :

$W=3\times 4\times 18$

W = 216 J

So, the work done by all the forces acting on it in time interval t = 0 to t = 3 s is 216 J. Hence, this is the required solution.