**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.

**Answer/Explanation**

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 60^{0} 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. [3]

**Answer/Explanation**

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.

**Answer/Explanation**

**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.

**Answer/Explanation**

**(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. [3]

**Answer/Explanation**

KE gain = ½ × 105 × (10^{2}– 5^{2})

WD against Resistance = 50 × 40

Total WD = 5937.5 J

**Alternative method**

10^{2}= 5^{2}+ 2 × 50 × *a* [*a* = 0.75]

DF – 40 = 105*a*

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.

**Answer/Explanation**

**(i)** PE gain is 32 000 J **(ii)** [KE gain = 1⁄2 160 × \(1.25^{2}\)]

KE gain is 125 J

**(iii)** WD by drum = 32 000 + 125 + 20 000

[P = 52 125 ÷ 41.7]

Power is 1250 W