Question

6) A 3 kg object is measured to move with velocity and acceleration given by the formulas 6 v(t) = A + Br, alt) = 3Br?, where the constants A = 7 m/s, and B = 0.5 m/s (a) What is the net force acting on the object as a function of the time variable ? ssign is b. (b) What is the instantaneous power exerted on the object as a function of 1? (c) What is the total work done on the object between the timest 0s and 1 38? Question to be Graded Manually.

Answer 1

Part A.

From Newton's 2nd law:

F_net = m*a

m = mass of object = 3 kg

a = acceleration of object = 3*B*t^2

Given that B = 0.5 m/s^4, So

a = 3*0.5*t^4 = 1.5*t^4

And

F = m*a = 3*(1.5*t^4)

F = 4.5*t^4

Part B. Instantaneous power is given by:

Power = Force*Velocity = F(t)*V(t)

V(t) = A + B*t^3 = (7 + 0.5*t^3)

So,

P(t) = 4.5*t^4*(7 + 0.5*t^3)

P(t) = 31.5*t^4 + 2.25*t^7

P(t) = 31.5t4 + 2.25t7

Part C.

Now Work-done is given by:

Power = dW/dt

dW = P(t)*dt

Integrating both side from t = 0 to t = 3 sec

| av = L*P) = út P(t) * dt (31.5* t4 +2.25 *t') * dt Remember: [ * + de = *1 + c W(t) = 131.5 * * + 2.25 * * * w() = (31.58 35.2.25 3* –(315 409 2.25 4 0%, W(t) = 3376.18125 J W(t) = 3376.2 J

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