Question

A time-varying horizontal force F(t) = At4 + Bt2 acts for 0.500 s on a 12.25-kg object, starting at time . In the SI system, A has the numerical value 4.50 and B has the numerical value 8.75. (a) What are the SI units of A and B? (b) What impulse does this force impart to the object? 5) (a) A: N/s4 = kg • m/s6, B: N/s2 = kg • m/s4 (b) 12.9 N • s, horizontally

Answer 1

Part a)

`A = (N)/(s^4)`

`B = (N)/(s^2)`

PART B)

`I = 0.393 Ns`

Step-by-step explanation:

PART A)

As we know that the force is given as

`F = At^4 + B t^2`

here we know that each term of the equation must have same dimensions

so we will have

`At^4 = N`

`A = (N)/(s^4)`

similarly for other term

`Bt^2 = N`

`B = (N)/(s^2)`

PART B)

Impulse given by the force is given as

`impulse = \int Fdt`

now we have

`I = \int (At^4 + Bt^2)dt`

`I = \int (4.50 t^4 + 8.75 t^2) dt`

`I = (4.50(0.5)^5)/(5) + (8.75(0.5)^3)/(3)`

`I = 0.028 + 0.36`

`I = 0.393 Ns`