Questions on drag and terminal speed

Question 1

A cyclist travels along a straight, level road. Name two resistive forces that act on the moving cyclist and state the direction in which each acts.

Accepted Answer

Air resistance (drag) on the cyclist and bicycle (1); rolling resistance at the tyres, or friction in the bearings (1). Both act backwards, opposite to the motion (1). Static friction from the road may act forwards, providing the traction that drives the bicycle, so it is not a resistive force here.

Question 2

Define terminal speed.

Accepted Answer

The constant maximum speed reached by a falling object when the resultant force on it is zero (1), so the drag, plus any upthrust, balances its weight and it stops accelerating (1).

Question 3

State how the drag force on an object depends on its speed, for motion through a fluid.

Accepted Answer

Drag increases as speed increases (1); at the speeds studied it rises roughly with the square of the speed (1).

Question 4

State what is meant by lift.

Accepted Answer

The component of the force exerted by a fluid on an object moving through it that acts perpendicular to the direction of flow. (1)

Question 5

A car travels at its maximum speed along a straight, level road. State the relationship between the driving force and the total resistive force, and state the car's acceleration.

Accepted Answer

The driving force equals the total resistive force (1); the resultant force is therefore zero, so the acceleration is zero. (1)

Question 6

State one similarity and one difference between the friction between two solid surfaces and the drag on an object moving through air.

Accepted Answer

Similarity: both are resistive forces that oppose (relative) motion (1). Difference: drag increases with speed, whereas friction between solid surfaces does not depend on speed (or: drag requires motion through a fluid). (1)

Question 7

A skydiver of weight 750 N falls at terminal speed. State the size and direction of the drag force on them, and explain your reasoning.

Accepted Answer

Drag = 750 N (1) Acting upwards (1) At terminal speed the resultant force is zero, so the upward drag exactly balances the downward weight (1)

Question 8

A small sphere falls at terminal speed through a liquid. Its weight is 0.30 N and the upthrust on it is 0.05 N. Calculate the drag force on the sphere, and explain why it is not accelerating.

Accepted Answer

At terminal speed: weight = upthrust + drag (1) Drag = 0.30 − 0.05 (1) Drag = 0.25 N (1) The three forces balance, giving zero resultant force and therefore zero acceleration (1)

Question 9

Describe the shape of the velocity–time graph for an object dropped from rest that reaches terminal speed, and explain the changing gradient.

Accepted Answer

The curve starts steep, with acceleration ≈ g (1), and its gradient decreases as drag grows with speed (1), levelling off to a horizontal line at the terminal speed where acceleration is zero (1).

Question 10

A raindrop of mass 65 mg falls vertically. At one instant the drag force on it is 4.2 × 10⁻⁴ N. Taking g = 9.81 m s⁻², calculate the acceleration of the raindrop at this instant.

Accepted Answer

Weight = mg = 6.5 × 10⁻⁵ × 9.81 = 6.4 × 10⁻⁴ N (1). Resultant = 6.4 × 10⁻⁴ − 4.2 × 10⁻⁴ = 2.2 × 10⁻⁴ N downwards (1). a = F/m = 3.3 m s⁻². (1)

Question 11

A skydiver can fall either spread out or head-first. Explain why the terminal speed is lower when the skydiver is spread out.

Accepted Answer

Spread out, the skydiver presents a larger area to the air, so at any given speed the drag is larger (1). The weight is unchanged, so the drag grows to equal the weight at a lower speed (1); at that speed the resultant force is zero, so the constant (terminal) speed reached is lower. (1)

Question 12

The engine of a car can deliver a maximum useful output power of 90 kW. At the car's top speed on a level road the total resistive force on it is 2.0 kN. (a) Calculate the top speed. (b) Explain why the car cannot travel faster than this speed.

Accepted Answer

(a) At top speed the driving force equals the resistive force, 2000 N (1); P = Fv, so v = P/F = 90 000/2000 = 45 m s⁻¹ (1). (b) At any higher speed the drag would exceed the greatest driving force the engine can supply (1), so the resultant force would act backwards and the car would decelerate back to the top speed. (1)

Question 13

A parachutist is falling at terminal speed when the parachute opens. Explain, in terms of the forces, why the speed decreases and how a new, lower terminal speed is reached.

Accepted Answer

Opening the parachute sharply increases the drag (1), so drag now exceeds weight and the resultant force acts upwards (1): the parachutist decelerates. As speed falls, drag falls (1), until drag again equals weight at a new, lower terminal speed (1).

Question 14

Assuming the drag force is proportional to the square of the speed, an object experiences a drag of 20 N at 5.0 m s⁻¹. Calculate the drag force at 10 m s⁻¹.

Accepted Answer

Drag ∝ v² (1) Doubling the speed multiplies drag by 4: D = 20 × (10/5.0)² (1) D = 80 N (1)

Question 15

Explain why a heavier skydiver of the same shape and size reaches a higher terminal speed than a lighter one.

Accepted Answer

Terminal speed is reached when drag equals weight (1). A heavier skydiver has a larger weight, so a larger drag is needed to balance it (1); with the same shape they must fall faster to generate that drag (1).

Question 16

A trainee parachutist of total weight 850 N must land at no more than 5.5 m s⁻¹. With a particular canopy open, the drag force at speed v is given by D = kv², where k = 24 kg m⁻¹. Deduce whether this canopy is suitable.

Accepted Answer

At the landing (terminal) speed, drag = weight (1): 24v² = 850, so v = √(850/24) (1) = 6.0 m s⁻¹ (1). This exceeds the 5.5 m s⁻¹ limit, so the canopy is not suitable. (1)

Question 17

A skydiver of mass 92 kg falls head-first with a terminal speed of 58 m s⁻¹. Assume the drag force is proportional to the square of the speed. Taking g = 9.81 m s⁻², determine the drag force and the acceleration of the skydiver at a speed of 29 m s⁻¹.

Accepted Answer

At terminal speed the drag equals the weight: 92 × 9.81 = 903 N (1). At half the terminal speed the drag is a quarter of this: 903/4 = 226 N (1). Resultant = 903 − 226 = 677 N downwards (1), so a = 677/92 = 7.4 m s⁻². (1)

Question 18

A cyclist starts from rest on a straight, level road and pedals so that the driving force stays constant. After some time she reaches her maximum speed, and she then stops pedalling. Air resistance increases with speed. Describe and explain, in terms of the forces acting, how her velocity changes throughout.

Accepted Answer

At the start the speed is low, so drag is small and the resultant force is almost the whole driving force: the acceleration is greatest here (1). As the speed rises the drag increases (1), so the resultant force, and therefore the acceleration, decreases (1). Maximum speed is reached when the drag (plus other resistance) has grown to equal the driving force: the resultant is zero and the velocity is constant (1). When she stops pedalling, only the resistive forces act, so there is a resultant force backwards and she decelerates (1). As the speed falls the drag falls, so the deceleration itself becomes smaller and the velocity dies away gradually. (1)

Question 19

Two spheres P and Q have the same diameter and the same surface finish, so the drag force on each is kv² with the same value of k. Each is released from rest and falls vertically through still air. Sphere Q has a mass of 0.050 kg. Take g = 9.81 m s⁻². Deduce the mass of sphere P and the value of k.

Accepted Answer

The terminal speed of each sphere is read where its curve becomes horizontal, giving 20 m s⁻¹ for P and 10 m s⁻¹ for Q (1)
At terminal speed the drag balances the weight, so mg = kv², and k is the same for both spheres, so the mass is proportional to the square of the terminal speed (1)
m_P = 0.050 × (20/10)² = 0.20 kg (1)
k = mg/v² = 0.050 × 9.81/10² = 4.9 × 10⁻³ kg m⁻¹, and P's data gives the same value (1)