Question

A particle is acted upon by a variable force such that the position of the particle of mass 2 kg varies with time as x = 4t2 + t, where x is in metre and t in second. The work done by the force in first 2 s is

Answer 1

Answer:

288 J

Explanation:

mass, m=2kg

time, t=2s

x=4t²+t

v=dx/dt = 8t+1

a=dv/dt = 8

Force, F=ma=2×8=16 N

x at 2s = 4×2²+2=18m

• Work done, W=F×x = 16×18 = 288J

Answer 2

Answer: The work done by the force in first 2 second is 288 J.

Explanation:

Mass of the particle = 2kg

The motion of the particle is represented by the equation:

x = 4t² + t ------------------------------(1)

Differentiating (1) we get

$\frac{d}{dx}$ x = $\frac{d}{dx}$ 4t² + t

=> v =   8t + 1 --------------------(2)

Differentiating (2) we get

$\frac{d}{dx}$ v = 8

=> a = 8 m/s²

Where x is in meter and t is in the second.

We know that work done = force x displacement

=> W = F x S ----------------------(3)

Where,

S = displacement

F = force

W = work done

Now,

x (t=0) = 0

x(t=2) = 4.2² + 2 = 18 m

Displacement of the particle in first 2 sec is: 18 - 0 =18

W = F . s

W = M a .s

W = 2 . 8 . 18 = 288 J

Therefore, The work done by the force in first 2 sec is 288J