A rubber band is stretched and then relaxed to its original length. The diagram shows the force-extension graph for this process.

Question 7

A rubber band is stretched and then relaxed to its original length. The diagram shows the force-extension graph for this process.

As the force is increased, the curve follows the path OPQ to extension e. As the force is reduced, the curve follows the path QRO to return to zero extension.

The area labelled X is between the curves OPQ and QRO. The area labelled Y is bounded by the curve QRO and the horizontal axis.

Which statement about the process is correct?

A Area X is the energy which heats the band as it is stretched to extension e.

B (Area X + area Y) is the minimum energy required to stretch the band to extension e.

C Area X is the elastic potential energy stored in the band when it is stretched to extension e.

D (Area Y – area X) is the net work done on the band during the process.

Reference: Past Exam Paper – November 2011 Paper 11 Q24 & P13 Q22 & June 2017 Paper 12 Q21

Solution:

Answer: B.

The work done in stretching the rubber band along the path OPQ is the area under the curve OPQ for loading. This corresponds to (area X + area Y). [B is correct]

Area X between the stretching (OPQ) the contraction (QRO) curves represents the amount of thermal energy (heat) dissipated (lost) in the rubber.

Area Y represents the recoverable elastic potential energy stored at maximum extension.

A stone of weight 3.0 N is fixed, using glue, to one end P of a rigid rod CP, as shown in Fig. 1.1.

Question 1

(a) Define the radian. [2]

(b) A stone of weight 3.0 N is fixed, using glue, to one end P of a rigid rod CP, as shown

in Fig. 1.1.

Fig. 1.1

The rod is rotated about end C so that the stone moves in a vertical circle of radius 85 cm.

The angular speed ω of the rod and stone is gradually increased from zero until the glue

snaps. The glue fixing the stone snaps when the tension in it is 18 N.

For the position of the stone at which the glue snaps,

(i) on the dotted circle of Fig. 1.1, mark with the letter S the position of the stone, [1]

(ii) calculate the angular speed ω of the stone. [4]

Reference: Past Exam Paper – June 2010 Paper 41 Q1

Solution:

(a) The radian is the angle (subtended) at the centre of a circle by an arc equal in length to the radius.

(b)

(i)

{The glue causes the stone to stay connected to the rod. So, it exerts a force towards the rod (since the rod is moving in circular motion, this force is always towards the centre.).

The centripetal force is the resultant force of the weight (always downwards) and the force (tension) exerted by the glue on the stone.

At the top, both the weight and the force act towards the centre. So, the centripetal force is largest (as it is due to both the weight and the force (due to W+F)).

Exactly to the right or left of C, the weight does not have any component towards or opposing the force exerted by the glue. So, the centripetal force is due to the forc exerted by the glue only (due to F only.)

For the glue to snap, the force opposing it should be sufficiently large.

(At the bottom,) when the stone is exactly below point C, its weight acts vertically downwards while the force exerted by the glue (to stay connected) is vertically up. The centripetal force is caused by the resultant of these 2 forces (due to F – W). It is in this position that the centripetal is the least and where the stone is most likely to snap.  

In any other position, the weight always acts downwards while the force by the glue is always towards the centre. So, in such positions, only a component of the weight will oppose the force due to the glue. Therefore, the force opposing the glue is maximum when the stone is exactly below C. It is the most likely position that the glue would snap.}

Point S is shown below C      

(ii)

{The centripetal acceleration is caused by the resultant force.

Resultant force = Tension – Weight (= Centripetal force)

Tension = Weight + Centripetal force}

(Maximum) Force (tension) = Weight of stone + Centripetal force

Centripetal force = mrω²

{Centripetal force = Tension – Weight = 18 – 3 = 15 N

W = mg                       So, mass m = W/g = 3.0 / 9.8

Radius r = 85 cm = 0.85 m}                                                  

15 = (3.0 / 9.8) × 0.85 × ω²

Angular speed ω = 7.6 rad s^-1            

A cyclist is travelling at a constant speed up a hill. The frictional force resisting the cyclist’s motion is 8.0 N.

Question 28

A cyclist is travelling at a constant speed up a hill. The frictional force resisting the cyclist’s motion is 8.0 N.

The cyclist uses 450 J of energy to travel a distance of 20 m.

What is the increase in the gravitational potential energy of the cyclist?

A 160 J                        B 290 J                        C 440 J                       D 610 J

Reference: Past Exam Paper – June 2017 Paper 12 Q15

Solution:

Answer: B.

Since the cyclist is moving at constant speed, he is NOT gaining kinetic energy.

From conservation of energy,

Work done by cyclist = Gain in GPE + Work done against friction

450 = Gain in GPE + (8.0×20)

450 = Gain in GPE + 160

 
Gain in GPE = 450 – 160 = 290 J