A horizontal force of 12N is applied to 1.5kg of block which rests on a horizontal surface. If the coefficient friction is 0.3,
1 answer:
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Answer:
I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s , I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s , I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s and I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s
Explanation:
The impulse is equal to the variation of the moment, to apply this relationship to our case, we will assume that initially the mouse was at rest
I = Δp = m -m v₀
I = m ( -v₀)
Bold indicates vector quantities, let's calculate the momentum of each mouse in for the x and y axes
We recommend bringing all units to the SI system
Mouse 1.
It has a mass of 22.3 g = 22.3 10⁻³ kg, a final velocity of (v = 0.349 i ^ - 0.301 j ^) m / s with an initial velocity of zero
Iₓ = m ( - v₀ₓ)
Iₓ = 22.3 10⁻³ (0.349 -0)
Iₓ = 7.78 10⁻³ J s
= m (
-
)
= 22.3 10⁻³ (-0.301)
= -6.71 10⁻³ J s
I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s
Mouse 2
Mass 17.9 g = 17.9 10⁻³ kg
Speed (-0.699 i ^ - 0.815 j ^) m / s
Iₓ = m ( - v₀ₓ)
Iₓ = 17.9 10⁻³ (-0.699 -0)
Iₓ = -12.5 10⁻³ J s
= 17.9 10⁻³ (-0.815 - 0)
= -14.6 10⁻³ J s
I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s
Mouse 3
Mass 19.1 g = 19.1 10⁻³ kg
Speed (0.745i ^ + 0.975 j ^) m / s
Iₓ = 19.1 10⁻³ (0.745 -0)
Iₓ = 14.2 10⁻³ J s
= 19.1 10⁻³(0.975 -0)
= 18.6 10⁻³ J s
I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s
Mouse 4
Mass 10.1 g = 10.1 10⁻³ kg
Speed (-0.905i ^ + 0.717j ^) m / s
Iₓ = 10.1 10⁻³ (-0.905 -0)
Iₓ = -9.14 10⁻³ J s
= 10.1 10⁻³ (0.717 -0)
= 7.24 10⁻³ J s
I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s