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andrey2020 [161]
2 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]2 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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bekas [8.4K]

Answer:

-54.12 V

Explanation:

The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.

W=\Delta E\\W=\Delta K+\Delta U\\W=K_f+\Delta U\\\Delta U=W-K_f\\\Delta U=6.10*10^{-4}J-1.50*10^{-4}J\\\Delta U=4.60*10^{-4}J

The change in electrostatic potential energy \Delta U, of one point charge q is defined as the product of the charge and the potential difference.

\Delta U=qV\\V=\frac{\Delta U}{q}\\V=\frac{4.60*10^{-4}J}{-8.50*10^{-6}C}\\V=-54.12 V

5 0
3 years ago
A 320 g air track cart traveling at 1.25 m/s collides with a stationary 270 g cart. What is the speed of the 270 g cart after th
Nutka1998 [239]

Answer:

The speed of the 270g cart after the collision is 0.68m/s

Explanation:

Mass of air track cart (m1) = 320g

Initial velocity (u1) = 1.25m/s

Mass of stationary cart (m2) = 270g

Velocity after collision (V) = m1u1/(m1+m2) = 320×1.25/(320+270) = 400/590 = 0.68m/s

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A 4.79 g bullet moving at 642.3 m/s penetrates a tree trunk to a depth of 4.35 cm. Use work and energy considerations to find th
vladimir1956 [14]

Answer:

Force, F=2.27\times 10^4\ N

Explanation:

Given that,

Mass of the bullet, m = 4.79 g = 0.00479 kg

Initial speed of the bullet, u = 642.3 m/s

Distance, d = 4.35 cm = 0.0435 m

To find,

The magnitude of force required to stop the bullet.

Solution,

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F.d=\dfrac{1}{2}m(v^2-u^2)

Finally, it stops, v = 0

F.d=-\dfrac{1}{2}m(u^2)

F=\dfrac{-mu^2}{2d}

F=\dfrac{-0.00479\times (642.3)^2}{2\times 0.0435}

F = -22713.92 N

F=2.27\times 10^4\ N

So, the magnitude of the force that stops the bullet is 2.27\times 10^4\ N

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