Question

A machine is driven by a 100 kg mass that falls 8.0m in 4.0s . It lifts a load oass 500kg vertically upwards.
Taking g = 10ms^-2 , calculate:
(a) the force exerted by the falling mass,
(b) the work done by the falling mass in its displacement by 8.0m,
(c) the power input to the machine.
(d) power output if power efficiency is 60% ​

Answer 1

Explanation:

a machine is driven by a 100 kg mass that Falls 8 metre in 4 second. it lifts a load of mass 500 kg vertically upward (i) what is the force in Newton exerted by falling mass (ii) what is the work done by 100 kg mass falling through 8m (iii)what is the power input of the machine if efficiency of machine is 75% (iv) what is the power output of the machine (v) what is the work done by the machine in 4 second

Answer 2

Given:–

• Mass = 100kg

• Displacement = 8.0m

• Load = 500kg

• Time = 4.0s

• g = 10ms^-2

To find:–

• Force exerted

• Work done

• Power input

• Power output if efficiency is 60%

As we know:–

$• \: \bold{force \: exerted \: by \: falling \: mass \: = \: mass \: \times g}$

$• \bold{work \: done \: by \: falling \: mass \: = force \: exerted \: by \: falling \: mass \: \times displacement}$

$• \: \bold{power \: input \: = \: \frac{work \: done \: on \: machine}{time} }$

$• \: \bold{efficiency \: = \: \frac{power \: output}{power \: input} }$

Step by step explaination:–

(a) As force exerted by falling mass = mass × g

So, mass is 100 kg and g is 10ms^-2.

That is,

→ 100 × 10

→ 1000 N

(b) As work done by the falling mass = force exerted × displacement .

That is,

→ mg × d

→ 1000 × 8

→ 8000J

(c) As power input = work done on machine/time.

That is,

$→ \dfrac{\cancel{8000 }}{\cancel4}$

→ 2000W

(d) Now we have to calculate the power output if efficiency is 60%.

As,

efficiency = power output/power input

Thus,

Power output = power input × efficiency

That is,

→ 2000 W = 0.6

→ 1200 W

Extra information:-

• A machine is a device by which we can either overcome a large resistive force(or load) at some point by applying a small force (or effort) at a convenient point and in a desired direction or by which we can obtain a gain in speed.

• The resistive or opposing force to be overcome by a machine is called the load.

• The force applied on the machine to overcome the load is called the effort.

• The ratio of the load to the effort is called the mechanical advantage of machine.

• The work done on the machine by the effort is called the work input.

• The work done on the machine by the on the load is called the load output.