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Oliga [24]
2 years ago
11

The figure above shows the net force exerted on an object as a function of the position of the object. The object starts from re

st at position x = 0 m and acquires a speed of 3.0 m / s after traveling the distance of 0.090 m shown above. What is the mass of the object?

Physics
1 answer:
weqwewe [10]2 years ago
5 0

Answer:

0.06 Kg

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Net Force (F) = 3 N

Mass (m) =?

Next, we shall determine the acceleration of the object. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Acceleration (a) =?

v² = u² + 2as

3² = 0² + (2 × a × 0.09)

9 = 0 + 0.18a

9 = 0.18a

Divide both side by 0.18

a = 9 / 0.18

a = 50 m/s²

Finally, we shall determine the mass of the object. This can be obtained as follow:

Net Force (F) = 3 N

Acceleration (a) = 50 N

Mass (m) =?

F = ma

3 = m × 50

Divide both side by 50

m = 3 / 50

m = 0.06 Kg

Therefore, the mass of the object is 0.06 Kg

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Actually Welcome to the Concept of the Projectile Motion.

Since, here given that, vertical velocity= 50m/s

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Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left
Dahasolnce [82]

Answer:

x_c= \dfrac{5}{9}L

I=\dfrac {7}{12}\lambda_ 0 L^3

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any  distance x from point A mass density

\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x

\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

dI=\lambda x^2dx

So total moment of inertia

I=\int_{0}^{L}\lambda x^2dx

By putting the values

I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx

By integrating above we can find that

I=\dfrac {7}{12}\lambda_ 0 L^3

Now to find location of center mass

x_c = \dfrac{\int xdm}{dm}

x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}

Now by integrating the above

x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}

x_c= \dfrac{5}{9}L

So mass moment of inertia I=\dfrac {7}{12}\lambda_ 0 L^3 and location of center of mass  x_c= \dfrac{5}{9}L

8 0
3 years ago
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