Question

A time-varying horizontal force F(t) = At4 + Bt2 acts for 0.500 s on a 12.25-kg object, starting attime t = 1.00 s. In the SI system, A has the numerical value 4.50 and B has the numerical value8.75. What impulse does this force impart to the object?

Answer 1

Answer:

3.82 Ns

Explanation:

Time varying horizontal Force is given as

F(t) = A t⁴ + B t²

F(t) = 4.50 t⁴ + 8.75 t²

Impulse imparted is given as

$I = \int_{0}^{t}Fdt$

$I = \int_{0}^{1}Fdt$

$I = \int_{0}^{1}(4.50 t^{4} + 8.75 t^{2})dt$

$I = ((0.9) (1)^{5} + (2.92) (1)^{3})$

$I = 3.82 N-s$

Answer 2

Answer:

impulse is 12.8614 kg

Explanation:

Given data

F(t) = At4 + Bt2

time t = 1.00 s

A = 4.50

B = 8.75

to find out

What impulse does this force impart to the object

solution

we know impulse is the change in momentum so we can right this as that

impulse I  = F i.e

dI = F(t) dt

we integrate it with limit 1 to 1.5

I = $\int_{1}^{1.5} At^4 + Bt^2$

I = $(At^5 / 5)^{1.5} _1$ + $(Bt^3 / 3)^{1.5} _1$

put the value A and B

I = (4.50(1.5)^5 / 5) - (4.50(1)^5 / 5)  + (8.75(1.5)^3 / 3) -  (8.75(1)^3 / 3)

I = (4.50(1.5)^5 / 5) - (4.50(1)^5 / 5)  + (8.75(1.5)^3 / 3) -  (8.75(1)^3 / 3)

I = 6.8343 - 0.9 + 9.8437 - 2.9166

impulse is 12.8614 kg