## Machines & inclined planes Questions

1.  An inclined plane of length 5m is used to raise a body of mass 60kg to the back of a lorry. If the

plane is inclined at an angle 25° from the horizontal, calculate the efficiency of the system given

that a constant force of 650 N is used to push the body up the plane

2.  Vicky performed an experiment using a pulley system as shown in the figure.

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(a) What is the V.R. of the system?

(b) Determine the M.A. of the system.  *

(c) Calculate the efficiency of the system.

(d) Explain why efficiency of a practical machine is always less than 100%

(e) If the load moves a distance of 5 cm. Find the work done on the load.

3.  The figure below shows a pulley system being used to raise a load. Use the information given

in the figure to answer questions (a) and (b)

(ii) If a load of 100N is raised by applying an effort of 48N, determine the efficiency of

the system.

4.  (a) (i) Define the term velocity ratio (V.R)

(ii) Name one machine that has a velocity ratio of less than one  (V.R < 1)

(b) The figure below shows a set-up used to find the mechanical advantage of a pulley system

On the axes provided sketch a graph of mechanical advantage (M.A) against load (L)

(c) A hydraulic machine is used to raise a load of 100kg at a constant velocity through a height

of 2.5m. The radius of the effort piston is 1.4cm while that of the load piston is 7.0cm. Given

that the machine is 80% efficient, calculate:-

(i) The effort needed

(ii) The energy wasted in using the machine

5.   (i) complete the diagram below to show how the pulley can be used to raise a load L by

applying an effort E

(ii) The pulley system above has a mechanical advantage of 3. Calculate the total work done

when a load of 60N is raised through a height of 9M

## Machines & inclined planes Answers

1.  M.A = 600M = 0.92307

650M

V.R = 1 = 2.366

Sin 25

= M.A = X 100

V.R

= 0.92307 X100

2.366 = 39.01%

2.  (a) V.R = 5

(b) MA = L

E

= 4000

1000

= 4

(c) eff. = M.A x 100%

VR

= 4/5

= 80%

(d) Some work is done overcoming friction or lifting the moving parts

(e) W = F x d

= 40,000 x 0.05 = 2000J

3.   VR = 4

A = L = 100

E 48

u = M.A x 100%

V.R

= 100 x ¼ = 52.08%

48

4.   (a) (i) Velocity ration is the distance moved by the effort to the distance moved by the load

in the same time

(ii) – Pulley belts

– Gears (any one)

(b) Graph

(c) (i) V.R = R2 = 7x 7 = 25

r2 1.4 x 1.4

Efficiency = M.A x 100%

V.R

M.A = r x V.R = 80 x 25 = 20

100

E = KL = 100 x 10 = 50N

M.A 20

(ii) EH = work output x 100%

Work input

Work output =mgh

= 100 x 10 x 2.5

= 2500J

80 = 2500 x 100

Work output

Work out put = 2500 x 100 = 3125J

80

Energy lost = 3125 – 2500

= 625J

5.  i)

ii) E   = L/M.A

= 60/3

= 20N

Total work done by effort

= E x Distance moved by effort

= 20 x 9 x V.R

= 20 x 9 x 4

= 720J

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