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user100 [1]
3 years ago
14

You put a 3 kg block in the box, so the total mass is now 9 kg, and you launch this heavier box with an initial speed of 5 m/s.

How long does it take to stop? Δt = s
Physics
1 answer:
OlgaM077 [116]3 years ago
8 0

Answer:

Δt=0.85 seconds

Explanation:

In this chase the speed does not change as the mass change.So we can use the follow equation to find the required time

Δt=Δv/gμ

To stop the final speed will be zero therefore the change in speed will be

Δv= vf-vi

Δv=0-5 m/s

Δv= -5 m/s

Now we plug our values for  Δv,g and μ to find time

Δt=Δv/gμ

Δt=(-5m/s) ÷(9.8m/s² × 0.6)

Δt=0.85 seconds

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Answer and Explanation:

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a=\frac{18.8}{3.6}

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(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².

Suppose positive referential is going up. To determine mass:

F_{r}=m.a

F_{t}-F_{g}=m.a

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Block m has 1.25 kg.

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F_{g}=1.25*9.8

F_{g}= 12.25 N

Comparing tension and weight:

\frac{12.25}{18.8} ≈ 0.65

We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.

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