U3 Videos FP 6.1.21

A toy car with a mass of 0.453 kg is moving up a semi-circular hill with a radius of 0.89 m. If its tangential velocity is 1.15 m/s when the hill makes an angle of 32° with the horizontal, what is the radius of the semi-circular hill?

• 0.453 m

• 1.15 m

• 0.89 m

• 32 m

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When analyzing the forces on the toy car moving up the semi-circular hill, the force of gravity is broken into components. Which of the following correctly describes these components relative to the car's motion?

• A component perpendicular to the hill's surface and a component parallel to the hill's surface.

• A component acting horizontally and a component acting vertically.

• A component acting towards the center of the Earth and a component acting away from the center of the Earth.

• A component acting in the direction of motion and a component acting opposite to the direction of motion.

Which of the following equations correctly represents the magnitude of the normal force (F_N) on the car at the given point on the semi-circular hill? (Where m is mass, g is acceleration due to gravity, θ is the angle of the hill with the horizontal, v_t is tangential velocity, and r is the radius of the hill.)

• F_N = mg sinθ + m(v_t^2/r)

• F_N = mg cosθ - m(v_t^2/r)

• F_N = mg cosθ + m(v_t^2/r)

• F_N = mg + m(v_t^2/r)

What is the magnitude of the force normal on the car, rounded to two significant digits, given the following values: mass = 0.453 kg, tangential velocity = 1.15 m/s, radius = 0.89 m, angle = 32°, and acceleration due to gravity = 9.81 m/s²?

• 3.8 N

• 4.4 N

• 5.1 N

• 2.9 N