# Physics 9702 Doubts | Help Page 37

__Question 228: [Units]__Spring constant k of coiled wire spring is given by equation

k = Gr

where r is radius of the wire, n is the number of turns of wire and R is
radius of each of the turns of wire. Quantity G depends on material from which
the wire is made.^{4}/ 4nR^{3}What is a suitable unit for G?

A N m

^{–2}B N m

^{–1}C N m D N m

^{2}

**Reference:**

*Past Exam Paper – November 2013 Paper 13 Q4*

__Solution 228:__**Answer: A.**

From Hooke’s law: F = kx

Spring constant, k = F / x Units of k: Nm

^{-1}

From the equation given,

G = 4nR

^{3}k / r

^{4}

Unit of R: m. Unit of r: m. n has no units.

Units of G: [m

^{3}] [Nm

^{-1}] [m

^{-4}] = Nm

^{-2}

__Question 229: [Work, Energy, Power]__**(a)**Define (i) work (ii) power [3]

**(b)**By reference to equations of motion, derive an expression for the kinetic energy E

_{k}of an object of mass m moving at speed v. [4]

**(c)**A car is travelling along a horizontal road with speed v, measured in metres per second. The power P, measured in watts, required to overcome external forces opposing the motion is given by the expression

P = cv + kv

^{3}

where c and k are constants.

(i) Use base units to obtain an SI unit for the constant k.

(ii) For one particular car, the numerical values, in SI units, of c and k are 240 and 0.98 respectively. Calculate the power required to enable the car to travel along a horizontal road at 31ms

^{-1}. [6]

**(d)**The car in (c) has mass 720kg. Using your answer to (c)(ii) where appropriate, calculate, for the car travelling at 31ms

^{-1},

(i) its kinetic energy

(ii) the magnitude of the external force opposing the motion of the car

(iii) the work done in overcoming the force in (ii) during a time of 5.0 minutes. [5]

**(e)**By reference to your answers to (d), suggest, with a reason, whether it would be worthwhile to develop a system whereby, when the car slows down, its kinetic energy would be stored for re-use when the car speeds up again. [2]

**Reference:**

*Past Exam Paper – J2000 / III / 2*

__Solution 229:__**(a)**

(i) The work done on a body is the product of the force applied in the body and the distance moved by the body in direction of the force.

(ii) Power is the rate of dong work.

**(b)**

For this part, check part (b) of Solution 204 at Physics 9702 Doubts | Help Page 33 -

*http://physics-ref.blogspot.com/2014/12/physics-9702-doubts-help-page-33.html*

**(c)**

(i)

SI base unit of Power P = kg ms

^{-2}m s

^{-1}= kgm

^{2}s

^{-3}

SI base unit of velocity v = ms

^{-1}

For homogeneity of the equation, the units on the right hand side should be equal to the units of P.

Units of kv

^{3}= kgm

^{2}s

^{-3}

Units of k = [kgm

^{2}s

^{-3}] [ms

^{-1}]

^{-3}= kgm

^{2}s

^{-3}m

^{-3}s

^{3}= kgm

^{-1}

(ii) Power P = 240 (31) + 0.98 (31

^{3}) = 36.6kW

**(d)**

(i) Kinetic energy = ½ mv

^{2}= 0.5 (720) (31

^{2}) = 3.46x10

^{5}J

(ii)

Power P = External force F x Velocity v

External force F = P / v = 366000 / 31 = 1.18kN

(iii)

Power = Work done / time

Work done = Power x time = 366000 x (5x60) = 1.10x10

^{7}J

**(e)**

Useful work = KE = 3.46x10

^{5}J

Work done in overcoming the external force opposing the motion of the car during a time of 5.0 minutes = 1.10x10

^{7}J

The useful work is comparatively very small. So, the efficiency of the system is low. So, it is not worthwhile to develop such a system.

__Question 230: [Forces > Equilibrium]__Diagram shows four forces applied to circular object.

Which of following describes resultant force and resultant torque on the object?

**Reference:**

*Past Exam Paper – June 2014 Paper 12 Q13*

__Solution 230:__**Answer: A.**

Vertical resultant force = 45 – 45 =
0

Horizontal resultant force = 30 + 30
= 60N to the right

The two 45N forces cause a clockwise
moment. The upper 30N force also causes a clockwise moment.

Only the lower 30N force causes an
anticlockwise moment.

So, the resultant torque is not
zero.

__Question 231: [Kinematics > Graph]__Graph shows how speed v of a sprinter changes with time t during a 100 m race.

What is the best estimate of the maximum acceleration of sprinter?

A 0.5 m s

^{–2}B 1 m s

^{–2}C 3 m s

^{–2}D 10 m s

^{–2}

**Reference:**

*Past Exam Paper – June 2014 Paper 13 Q9*

__Solution 231:__**Answer: C.**

Acceleration is represented by the gradient
of the velocity-time graph.

Gradient is zero when the
graph is a horizontal line and infinite when the graph is a vertical line. Neglecting the + or - sign that would indicate direction for the
vectors, it can be concluded a line which is closer to being horizontal will
have a small value of gradient and a line which is closer to being vertical
will have a large value of gradient.

The section between time t = 2 and time
t = 4 is closer to a vertical line, so its gradient is the highest among the
lines present in the graph.

Considering the following estimated
points: (2, 1) and (4, 8),

Gradient = (8 – 1) / (4 – 2) = 3.5ms

^{-2}
This is closest to the choice C.

__Question 232: [Forces]__Three coplanar forces, each of magnitude 10 N, act through same point of a body in directions shown.

What is the magnitude of resultant force?

A 0 N B 1.3 N C 7.3 N D 10 N

**Reference:**

*Past Exam Paper – November 2006 Paper 1 Q15*

__Solution 232:__**Answer: C.**

Each of the 2 diagonal forces will have a horizontal component of 10sin30 N in opposite directions. So, it can be seen the resultant horizontal force is zero.

Sum of upward vertical components of forces = 10cos30 + 10cos30 = 17.3N

Resultant force = 17.3 – 10 = 7.3N

__Question 233: [Kinematics > Linear motion > Graph]__Graph of velocity against time for an object moving in straight line is shown.

What is corresponding graph of displacement against time?

**Reference:**

*Past Exam Paper – November 2012 Paper 12 Q9*

__Solution 233:__**Answer: C.**

The gradient of the velocity-time graph gives the acceleration and the area under the graph gives the displacement.

Since the velocity is always positive, the displacement is also always positive [B is incorrect] and keeps on increasing [D is incorrect].

The gradient of the displacement-time graph gives the velocity.

For the first section, the velocity

__increases__at a constant rate with time. This causes the displacement to increase with time, with an increasing gradient (as velocity is represented by the gradient) since with the increasing velocity, the displacement (in each second) becomes greater for every second. [D is incorrect since the gradient of the displacement-time graph is constant.]

For the middle section, the velocity is constant, so the graph of displacement has a constant gradient for the corresponding section. For the last section, the velocity decreases with time, so the gradient of displacement-time graph decreases with time.

Ref to solution 229 (c-i) sir can you please let me know why you are not considering the factor cv to find the base units?

ReplyDeleteSince there is a '+', cv and kv^3 both have the same units independently, which are both equal to the unit of Power P. We need to find the unit of k, so there's no need to consider the cv part.

Deleteq14 june 2003 paper 1

ReplyDeletethank u

See solution 938 at

Deletehttp://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-193.html