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?

Question

05-08-2019
Physics

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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 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?

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JemdetNasr JemdetNasr · Answer 1 · 2019-08-08T07:42:11+02:00

JemdetNasr JemdetNasr

08-08-2019

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

[tex]I = \int_{0}^{t}Fdt[/tex]

[tex]I = \int_{0}^{1}Fdt[/tex]

[tex]I = \int_{0}^{1}(4.50 t^{4} + 8.75 t^{2})dt[/tex]

[tex]I = ((0.9) (1)^{5} + (2.92) (1)^{3})[/tex]

[tex]I = 3.82 N-s[/tex]

Answer Link

DeniceSandidge DeniceSandidge · Answer 2 · 2019-08-08T07:47:12+02:00

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 = [tex]\int_{1}^{1.5} At^4 + Bt^2[/tex]

I = [tex](At^5 / 5)^{1.5} _1[/tex] + [tex](Bt^3 / 3)^{1.5} _1[/tex]

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 = 6.8343 - 0.9 + 9.8437 - 2.9166

impulse is 12.8614 kg