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A small car with mass m and speed 2v and a large car with mass 2m and speed v both travel the same circular...

A small car with mass m and speed 2v and a large car with mass 2m and speed v both travel the same circular section of an unbanked road. If the frictional force required to keep the small car on the road without skidding is F, then the frictional force required to keep the large car on the road without skidding is
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F
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2F
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F/4
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4F
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F/2

Answer

To determine the frictional force required to keep both cars on the circular path without skidding, we start by analyzing the forces acting on each car.
### For the Small Car:
1. **Mass**: \( m \)
2. **Speed**: \( 2v \)
3. **Centripetal Force Required**: The centripetal force required to keep the car moving in a circle of radius \( r \) is given by the formula:
\[
F_c = \frac{mv^2}{r}
\]
Since the speed of the small car is \( 2v \), we substitute this into the equation:
\[
F_c = \frac{m(2v)^2}{r} = \frac{m \cdot 4v^2}{r} = \frac{4mv^2}{r}
\]
4. **Frictional Force**: The frictional force \( F \) is providing this centripetal force, hence:
\[
F = \frac{4mv^2}{r}
\]
### For the Large Car:
1. **Mass**: \( 2m \)
2. **Speed**: \( v \)
3. **Centripetal Force Required**: For the large car moving at speed \( v \):
\[
F_{c, large} = \frac{(2m)v^2}{r} = \frac{2mv^2}{r}
\]
### Comparing Frictional Forces:
Now, we compare the frictional forces required for both cars to maintain their circular motion without skidding:
- For the small car, the frictional force is:
\[
F = \frac{4mv^2}{r}
\]
- For the large car, the frictional force required is:
\[
F_{large} = \frac{2mv^2}{r}
\]
### Relation Between Forces:
To express \( F_{large} \) in terms of \( F \):
\[
F = \frac{4mv^2}{r} \quad \text{(small car)}
\]
\[
F_{large} = \frac{2mv^2}{r}
\]
We can express \( F_{large} \) in terms of \( F \):
\[
F_{large} = \frac{2mv^2}{r} = \frac{1}{2} \cdot \frac{4mv^2}{r} = \frac{F}{2}
\]
### Conclusion:
Thus, the frictional force required to keep the large car on the road without skidding is:
F/2

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