> #mechanics , #physics > A 0.5kg stone is whirled on the end of a 1.2m string in a vertical circle at a constant speed of 6 m/s.Calculate the net force acting on the stone at a) the highest point b)lowest point?
###### Anonymous
more than 3 weeks ago

# A 0.5kg stone is whirled on the end of a 1.2m string in a vertical circle at a constant speed of 6 m/s.Calculate the net force acting on the stone at a) the highest point b)lowest point?

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more than 3 weeks ago
a) At the highest point of the circle, the stone is moving in a circular path with a constant speed. The net force acting on the stone at this point is equal to the centripetal force required to keep the stone moving in a circle.

The centripetal force is given by the formula:

F = (m * v^2) / r

Where:
F = centripetal force
m = mass of the stone = 0.5 kg
v = speed of the stone = 6 m/s
r = radius of the circle = 1.2 m

Plugging in the values, we get:

F = (0.5 * 6^2) / 1.2
F = (0.5 * 36) / 1.2
F = 18 / 1.2
F = 15 N

Therefore, the net force acting on the stone at the highest point is 15 N.

b) At the lowest point of the circle, the net force acting on the stone is the sum of the centripetal force and the force due to gravity.

The centripetal force is still given by the formula:

F = (m * v^2) / r
F = (0.5 * 6^2) / 1.2
F = 15 N

The force due to gravity is given by:

Fg = m * g
Fg = 0.5 * 9.8
Fg = 4.9 N

Therefore, the net force acting on the stone at the lowest point is:

Fnet = F + Fg
Fnet = 15 + 4.9
Fnet = 19.9 N

Therefore, the net force acting on the stone at the lowest point is 19.9 N.
0
more than 3 weeks ago
a) Highest point

At the highest point, the stone is moving at a constant speed in a circular path. The net force acting on the stone is the centripetal force, which is provided by the tension in the string.

The centripetal force is given by:

```
Fc = mv^2/r
```

where:

- Fc is the centripetal force
- m is the mass of the stone
- v is the speed of the stone
- r is the radius of the circle

Plugging in the given values, we get:

```
Fc = (0.5 kg)(6 m/s)^2 / (1.2 m)
Fc = 15 N
```

Therefore, the net force acting on the stone at the highest point is 15 N.

b) Lowest point

At the lowest point, the stone is still moving at a constant speed in a circular path, but the direction of the centripetal force has changed. Now, the centripetal force is provided by the tension in the string and the force of gravity acting on the stone.

The net force acting on the stone is the sum of the centripetal force and the force of gravity:

```
Fnet = Fc + mg
```

where:

- Fnet is the net force
- Fc is the centripetal force
- mg is the force of gravity

The centripetal force is still given by:

```
Fc = mv^2/r
```

The force of gravity is given by:

```
mg = m - g
```

where:

- g is the acceleration due to gravity (9.8 m/s^2)

Plugging in the given values, we get:

```
Fnet = 15 N + (0.5 kg)(9.8 m/s^2)
Fnet = 24.9 N
```

Therefore, the net force acting on the stone at the lowest point is 24.9 N.
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