Questions on density and Hooke's law

Question 1

Define density and state its SI unit.

Accepted Answer

Density is mass per unit volume, ρ = m/V (1). SI unit: kilogram per cubic metre, kg m⁻³ (1).

Question 2

A block has a mass of 540 g and a volume of 200 cm³. Calculate its density in g cm⁻³ and in kg m⁻³.

Accepted Answer

ρ = m/V = 540/200 = 2.7 g cm⁻³ (1) = 2700 kg m⁻³ (1)

Question 3

State Hooke's law for a spring.

Accepted Answer

The extension of a spring is directly proportional to the force applied (1), up to the limit of proportionality: F = kΔL (1).

Question 4

Distinguish between the limit of proportionality and the elastic limit of a wire.

Accepted Answer

The limit of proportionality is the point beyond which force is no longer proportional to extension, where the force–extension graph stops being straight (1). The elastic limit is the point beyond which the wire no longer returns to its original length when the force is removed (1).

Question 5

State how the energy stored in a stretched cord is found from its force–extension graph when the graph is curved.

Accepted Answer

The stored energy is the area under the force–extension graph, found by counting squares (½FΔL only applies while the graph is straight) (1).

Question 6

A spring extends by 4.0 cm when a force of 2.0 N is applied. Calculate (a) its spring constant and (b) the force needed to extend it by 6.0 cm (within its limit of proportionality).

Accepted Answer

(a) k = F/ΔL = 2.0/0.040 (1) k = 50 N m⁻¹ (1) (b) F = kΔL = 50 × 0.060 = 3.0 N (1)

Question 7

Steel has a density of 7800 kg m⁻³. Express this in g cm⁻³ and calculate the mass of a 50 cm³ steel component.

Accepted Answer

7800 kg m⁻³ = 7.8 g cm⁻³ (1) m = ρV = 7.8 × 50 (1) m = 390 g = 0.39 kg (1)

Question 8

A spring of spring constant 200 N m⁻¹ is stretched by 0.10 m within its limit of proportionality. Calculate the energy stored, and confirm your value using ½FΔL.

Accepted Answer

E = ½kΔL² = ½ × 200 × 0.10² (1) E = 1.0 J (1) Check: F = kΔL = 20 N (1) ½FΔL = ½ × 20 × 0.10 = 1.0 J (1)

Question 9

An irregular metal object of mass 0.50 kg is lowered into a measuring cylinder and displaces 62.5 cm³ of water. Calculate the density of the metal in kg m⁻³.

Accepted Answer

V = 62.5 cm³ = 62.5 × 10⁻⁶ m³ (1) ρ = m/V = 0.50/(62.5 × 10⁻⁶) (1) ρ = 8000 kg m⁻³ (1)

Question 10

A foundry casts a brass component by mixing 60 cm³ of copper (density 8900 kg m⁻³) with 40 cm³ of zinc (density 7100 kg m⁻³). Assuming the total volume is unchanged on mixing, calculate the density of the brass.

Accepted Answer

Mass of copper = 8900 × 60 × 10⁻⁶ = 0.534 kg (1). Mass of zinc = 7100 × 40 × 10⁻⁶ = 0.284 kg (1). Total mass 0.818 kg in a volume of 100 × 10⁻⁶ m³ (1), so ρ = 8180 kg m⁻³ (1).

Question 11

A metal cube of side 3.0 cm has a mass of 216 g. Use the table to determine which metal the cube is most likely made of.
aluminium 2700 kg m⁻³; titanium 4500 kg m⁻³; steel 8000 kg m⁻³; lead 11300 kg m⁻³

Accepted Answer

V = 3.0³ = 27 cm³ (1). ρ = 216/27 = 8.0 g cm⁻³ = 8000 kg m⁻³ (1). This matches steel (1).

Question 12

The force–extension graph for a rubber cord is curved. A student counts 34 whole squares under the curve up to an extension of 0.30 m, on a grid where each square represents 0.5 N by 0.02 m. Calculate the energy stored, and explain why the student counted squares rather than using ½FΔL.

Accepted Answer

Each square represents 0.5 × 0.02 = 0.01 J (1). Energy = 34 × 0.01 = 0.34 J (1). ½FΔL is the area of a triangle and only applies when the graph is a straight line through the origin; for a curved graph the area must be found directly (1).

Question 13

Two springs have spring constants 50 N m⁻¹ and 100 N m⁻¹. Calculate the effective spring constant when they are combined (a) in series and (b) in parallel, and find the extension of each combination under a 3.0 N load.

Accepted Answer

(a) Series: 1/k = 1/50 + 1/100, so k = 33.3 N m⁻¹ (1) Extension = 3.0/33.3 = 0.090 m (1) (b) Parallel: k = 50 + 100 = 150 N m⁻¹ (1) Extension = 3.0/150 = 0.020 m (1)

Question 14

A spring of spring constant 800 N m⁻¹ is compressed by 5.0 cm and used to launch a 0.20 kg ball horizontally. Assuming all the stored energy is transferred to the ball, calculate the launch speed.

Accepted Answer

Stored energy = ½kΔL² = ½ × 800 × 0.050² (1) = 1.0 J (1) ½mv² = 1.0, so v = √(2 × 1.0/0.20) (1) v = 3.16 m s⁻¹ (1)

Question 15

Explain what happens to a spring stretched beyond its limit of proportionality, and how this appears on a force–extension graph.

Accepted Answer

Beyond the limit of proportionality the extension is no longer proportional to the force (1), so the graph stops being a straight line and curves (1). Past the elastic limit the spring no longer returns to its original length when unloaded, which is plastic behaviour (1).

Question 16

A jumping toy contains a spring of spring constant 450 N m⁻¹ that is compressed by 3.0 cm before release. The toy has a mass of 12 g. Assuming all the stored energy is transferred to gravitational potential energy, calculate the maximum height the toy reaches. (g = 9.81 m s⁻².)

Accepted Answer

Convert to SI: ΔL = 0.030 m, m = 0.012 kg (1). Stored energy = ½kΔL² = ½ × 450 × 0.030² = 0.20 J (1). Equate to mgh: 0.20 = 0.012 × 9.81 × h (1), so h = 1.7 m (1).

Question 17

A supplier advertises springs with a spring constant of 25.0 N m⁻¹, accurate to within 5%. A student hangs a load of 6.6 N from one of the springs and measures its extension as 0.240 m, within the limit of proportionality. Deduce whether this spring meets the advertised claim.

Accepted Answer

k = F/ΔL = 6.6/0.240 = 27.5 N m⁻¹ (1). 5% of 25.0 is 1.25, so the claim allows 23.75 to 26.25 N m⁻¹ (1). 27.5 N m⁻¹ lies above the upper bound (1), so the spring is stiffer than advertised and does not meet the claim (1).

Question 18

A copper wire is loaded steadily until it is stretched beyond its elastic limit, and the load is then removed. Describe fully the behaviour of the wire during loading and unloading, with reference to its force–extension graph and to the energy transfers involved.

Accepted Answer

During loading the graph is a straight line through the origin, obeying Hooke's law, up to the limit of proportionality (1); beyond this the graph curves, and past the elastic limit the deformation becomes plastic (permanent) (1). On unloading, the graph is a straight line parallel to the original loading line (1), meeting the extension axis at a non-zero value, the permanent extension (1). The work done during loading exceeds the energy recovered during unloading, the difference being the area between the loading and unloading lines (1); this energy has gone into rearranging the internal structure of the metal and heating it, and is not recovered (1).

Question 19

A metal wire is loaded steadily until it has been stretched well beyond its elastic limit, and the load is then removed steadily. The graph shows the force against extension for the whole loading and unloading. Determine the permanent extension of the wire, the work done on the wire during loading, and the energy that is not recovered when the load is removed.

Accepted Answer

The permanent extension is read where the unloading line meets the extension axis, giving 7.0 mm (1)
The work done during loading is the area under the loading path, a triangle up to the corner plus a trapezium beyond it (1)
½ × 4.0 × 10⁻³ × 24 + ½ × (24 + 30) × 8.0 × 10⁻³ = 0.048 + 0.216 = 0.264 J (1)
The energy recovered is the area under the unloading line, a triangle of base (12.0 − 7.0) mm and height 30 N, that is ½ × 5.0 × 10⁻³ × 30 = 0.075 J (1)
The energy not recovered is 0.264 − 0.075 = 0.19 J, the area of the loop enclosed between the two paths, which has gone into rearranging the metal's structure and warming it (1)