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andrey2020 [161]
3 years ago
11

A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal

l moves so fast that the string is always taut and perpendicular to the velocity of the ball. As the ball swings from its lowest point to its highest point, what is the work done by gravity and by the tension?
Physics
1 answer:
Brut [27]3 years ago
6 0

Answer:

  • The gravity does a work of - 117.6 Joules.
  • The tension does not do work as the force is perpendicular to the direction of motion at any point in the trajectory.

Explanation:

The work done by the gravity simply is the difference in gravitational potential energy multiplied by -1:

W_g = - \Delta E_p = - (mgh_f  - m g h_i)

where m is the mass of the ball, g is the acceleration due to gravity, h_f is the final height and h_i is the initial height.

So, if the radius is 2.00 m, then the difference of height will be 4 meters:

W_g = - mg (h_f - h_i)

W_g = - 3.00 \ kg \ 9.8 \frac{m}{s^2} \ 4 \m

W_g = - 117.6 Joules

As the tension is perpendicular to the velocity of the ball, the force is always perpendicular to the direction of motion. So, the differential of work will be:

dW = \vec{F}  d\vec{r} = 0

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Answer:

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a) The pressure of the front tires is 22 psi. The appropriate conversion factor is 1 psi = 6.89 kPa. The pressure of the tires in kPa is:

22 psi × (6.89 kPa / 1 psi) = 1.5 × 10² kPa

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A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi
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Answer:

1) W_{object} = 400 N

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

\sum F_{Tot} = 0

In this case we just have y-direction forces.

\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0

Now, let's solve the equation for W(object).

W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N

2 ) To find the position of the heaviest weight we need to use the torque definition.

\sum \tau = 0

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

L = 1.50 m  ; total length of the bar.

D_{AP} = x  ; distance between Tension A and P point.

D_{BP} = L-x ; distance between Tension B and P point.

D_{W_{bar}P} = \frac{L}{2}-x ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0

Finally we just need to solve it for x.

x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}

x = 0.28 m

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

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