Question
bus of mass 2500 kg goes around corner of radius 50m at velocity of 5m/s what is the force needed for the bus to go around the corner?
Answers
GPT-4o mini
To determine the force needed for the bus to go around the corner, we need to calculate the centripetal force required to keep the bus moving in a circular path. The formula for centripetal force (\( F_c \)) is given by:
\[
F_c = \frac{mv^2}{r}
\]
where:
- \( m \) is the mass of the bus (2500 kg),
- \( v \) is the velocity of the bus (5 m/s),
- \( r \) is the radius of the corner (50 m).
Now, substituting the values into the formula:
\[
F_c = \frac{2500 \, \text{kg} \times (5 \, \text{m/s})^2}{50 \, \text{m}}
\]
Calculating \( (5 \, \text{m/s})^2 \):
\[
(5 \, \text{m/s})^2 = 25 \, \text{m}^2/\text{s}^2
\]
Now substitute back:
\[
F_c = \frac{2500 \, \text{kg} \times 25 \, \text{m}^2/\text{s}^2}{50 \, \text{m}}
\]
\[
F_c = \frac{62500 \, \text{kg} \cdot \text{m}^2/\text{s}^2}{50 \, \text{m}}
\]
Calculating this gives:
\[
F_c = 1250 \, \text{N}
\]
Therefore, the centripetal force needed for the bus to go around the corner is **1250 N**.
\[
F_c = \frac{mv^2}{r}
\]
where:
- \( m \) is the mass of the bus (2500 kg),
- \( v \) is the velocity of the bus (5 m/s),
- \( r \) is the radius of the corner (50 m).
Now, substituting the values into the formula:
\[
F_c = \frac{2500 \, \text{kg} \times (5 \, \text{m/s})^2}{50 \, \text{m}}
\]
Calculating \( (5 \, \text{m/s})^2 \):
\[
(5 \, \text{m/s})^2 = 25 \, \text{m}^2/\text{s}^2
\]
Now substitute back:
\[
F_c = \frac{2500 \, \text{kg} \times 25 \, \text{m}^2/\text{s}^2}{50 \, \text{m}}
\]
\[
F_c = \frac{62500 \, \text{kg} \cdot \text{m}^2/\text{s}^2}{50 \, \text{m}}
\]
Calculating this gives:
\[
F_c = 1250 \, \text{N}
\]
Therefore, the centripetal force needed for the bus to go around the corner is **1250 N**.