# CIE A level Math -Mechanics : 4.5 Energy, work and power: principle of conservation of energy: Exam Style Questions Paper 4

### Question

A particle of mass 0.6 kg is dropped from a height of 8 m above the ground. The speed of the particle at the instant before hitting the ground is 10 $$m s^{−1}$$ . Find the work done against air resistance.

PE loss = 0.6 × 10 × 8 [= 48]
KE gain =$$1⁄2 (0.6) 10^{ 2}$$ [= 30]
WD against Res = 48 – 30 = 18 J

### Question

A winch operates by means of a force applied by a rope. The winch is used to pull a load of mass  50 kg up a line of greatest slope of a plane inclined at 600 to the horizontal. The winch pulls the load
a distance of 5 m up the plane at constant speed. There is a constant resistance to motion of 100 N.
Find the work done by the winch.                                                                                                                                      

Force exerted by winch =50 sin 60 + 100 [= 433.0 + 100 533.0 ]

Work done =5× (50g sin 60 + 100)

Work done = 2670 J

Alternative method :

PE increase 50×  5sin 60

Work done = 50g × 5sin 60 + 100 × 5

Work done = 2670 J

### Question

A particle of mass 1.3 kg rests on a rough plane inclined at an angle 1 to the horizontal, where $$tan\Theta =\frac{12}{5}$$.The coefficient of friction between the particle and the plane is $$\mu$$.

(i) A force of magnitude 20N parallel to a line of greatest slope of the plane is applied to the particle
and the particle is on the point of moving up the plane. Show that $$\mu$$=1.6.The force of magnitude 20 N is now removed.
(ii) Find the acceleration of the particle.

(iii) Find the work done against friction during the first 2 s of motion.

4(i)
R = 13 cos 67.4 = 13 (5/13) [R = 5]F+ 13 sin 67.4 =F + 13(12/13) = 20 [F = 8]

μ = 8/5 = 1.6

4(ii) 13 sin 67.4 –F = 1.3a F=μR = 8→[4 = 1.3
a] a = 3.08$$s^{-2}$$

4(iii) s = 0+ 0.5× $$(40/13)\times 2^{2}$$[=80/13=6.15]

WD = 8× 6.15 WD = 49.2 J

s = 0+ 0.5× (40/13)× $$2^{2}$$ [= 80/13 = 6.15]

[v = (40/13) × 2]and [WD = 1.3g(80/13)(12/13) – 1⁄2× 1.3
× (80/13)(12/13)] WD = 49.2 J WD = 49.2 J

### Question

A block of mass 25 kg is pulled along horizontal ground by a force of magnitude 50 N inclined at 10Å above the horizontal. The block starts from rest and travels a distance of 20 m. There is a constant
resistance force of magnitude 30 N opposing motion.
(i) Find the work done by the pulling force.
(ii) Use an energy method to find the speed of the block when it has moved a distance of 20 m.
6 A block of mass 25 kg is pulled along horizontal ground by a force of magnitude 50 N inclined at 10Å
above the horizontal. The block starts from rest and travels a distance of 20 m. There is a constant
resistance force of magnitude 30 N opposing motion.
(i) Find the work done by the pulling force.
(ii) Use an energy method to find the speed of the block when it has moved a distance of 20 m.
(iii) Find the greatest power exerted by the 50 N force.Find the greatest power exerted by the 50 N force. After the block has travelled the 20 m, it comes to a plane inclined at 5Å to the horizontal. The force of 50 N is now inclined at an angle of 10Å to the plane and pulls the block directly up the plane (see diagram). The resistance force remains 30 N.
(iv) Find the time it takes for the block to come to rest from the instant when it reaches the foot of the inclined plane.

(i) [Work done = 50 cos10 × 20] = 984.8 J

(ii)$$[984.8 = 1⁄2\times 25v^{2}+30\times 20]$$ v = $$5.55ms^{-1}$$

(iii) Max power = 50cos10 × 5.55 = 273W

(iv) [50 cos10 – 30 – 25g sin5 = 25a] a =$$– 0.102ms^{-2}$$
[0 = 5.55 – 0.102t]

Time t = 54.4 s

50 cos 10 × s + 1⁄2 × 25 × $$5.55^{2}$$= 25g × s sin 5 + 30 × s

t = 302/5.55 = 54.4 s

### Question

A cyclist has mass 85 kg and rides a bicycle of mass 20 kg. The cyclist rides along a horizontal road against a total resistance force of 40 N. Find the total work done by the cyclist in increasing his speed from 5 m s−1 to 10 m s−1 while travelling a distance of 50 m. 

KE gain = ½ × 105 × (102– 52)

WD against Resistance = 50 × 40

Total WD = 5937.5 J

Alternative method

102= 52+ 2 × 50 × a [a = 0.75]

DF – 40 = 105a

DF = 40 + 105 × 0.75 = 118.75

Total WD = 118.75 × 50 = 5937.5 J

### Question A load of mass 160 kg is pulled vertically upwards, from rest at a fixed point O on the ground, using a winding drum. The load passes through a point A, 20 m above O, with a speed of $$1.25 m s^{−1}$$(see diagram). Find, for the motion from O to A,
(i) the gain in the potential energy of the load,
(ii) the gain in the kinetic energy of the load.  The power output of the winding drum is constant while the load is in motion.
(iii) Given that the work done against the resistance to motion from O to A is 20 kJ and that the time taken for the load to travel from O to A is 41.7 s, find the power output of the winding drum.

(ii) [KE gain = 1⁄2 160 × $$1.25^{2}$$]