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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

Respuesta :

Answer:

Part a)

[tex]A = \frac{N}{s^4}[/tex]

[tex]B = \frac{N}{s^2}[/tex]

PART B)

[tex]I = 0.393 Ns[/tex]

Explanation:

PART A)

As we know that the force is given as

[tex]F = At^4 + B t^2[/tex]

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

so we will have

[tex]At^4 = N[/tex]

[tex]A = \frac{N}{s^4}[/tex]

similarly for other term

[tex]Bt^2 = N[/tex]

[tex]B = \frac{N}{s^2}[/tex]

PART B)

Impulse given by the force is given as

[tex]impulse = \int Fdt[/tex]

now we have

[tex]I = \int (At^4 + Bt^2)dt[/tex]

[tex]I = \int (4.50 t^4 + 8.75 t^2) dt[/tex]

[tex]I = \frac{4.50(0.5)^5}{5} + \frac{8.75(0.5)^3}{3}[/tex]

[tex]I = 0.028 + 0.36[/tex]

[tex]I = 0.393 Ns[/tex]

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