You are riding in a school bus. As the bus rounds a flat curve at constant speed, a lunch box with a mass of 0.470 suspended from the ceiling of the bus by a string of length 1.85 is found to hang at rest relative to the bus when the string makes an angle of 29.0 with the vertical. In this position the lunch box is a distance 52.0 from the center of curvature of the curve.
What is the speed of the bus?
Domensions must be provided with numbers.
s=m
To find the speed of the bus, we need to consider the forces acting on the lunch box and use them to determine the centripetal force.
1. Draw a free-body diagram for the lunch box. In this case, there are two forces acting on the lunch box: tension (T) and weight (mg).
2. Decompose the weight (mg) into its components. The weight can be split into two forces: the vertical component (mgcosθ) and the horizontal component (mgsinθ), where θ is the angle the string makes with the vertical.
3. Identify the net force in the horizontal direction. The only force causing the lunch box to move in the horizontal direction is the horizontal component of tension (Tcosθ). So, the net force (Fnet) in the horizontal direction is equal to Tcosθ.
4. Determine the centripetal force. Since the lunch box is moving in a curved path, there must be a centripetal force acting on it. The centripetal force (Fc) is given by the equation Fc = (m * v²) / r, where m is the mass, v is the velocity, and r is the radius of the curved path.
5. Equate the net force to the centripetal force. Set the equation Fnet = Fc, which gives us Tcosθ = (m * v²) / r.
6. Solve for the speed of the bus. Rearrange the equation to solve for v: v = √[(T * r * cosθ) / m].
7. Plug in the given values. Substitute the values into the equation: T = 0.470 kg (mass of the lunch box), r = 52.0 m (distance from the center of curvature of the curve), and θ = 29.0°.
8. Calculate the speed of the bus. Using a calculator, evaluate the equation to find the speed of the bus.
Following these steps, you will be able to find the speed of the bus.