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

Answer:

Part a)

$A = \frac{N}{s^4}$

$B = \frac{N}{s^2}$

PART B)

$I = 0.393 Ns$

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

$At^4 = N$

$A = \frac{N}{s^4}$

similarly for other term

$Bt^2 = N$

$B = \frac{N}{s^2}$

PART B)

Impulse given by the force is given as

$impulse = \int Fdt$

now we have

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

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

$I = \frac{4.50(0.5)^5}{5} + \frac{8.75(0.5)^3}{3}$

$I = 0.028 + 0.36$

$I = 0.393 Ns$