A pulley system with a velocity ratio of 4 is used to lift a load of 100kgf through a vertical height is 15m. The effort required to do so is 40kgf which is applied in the downward direction. calculate Workdone by the effort

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Solution

Given
Velocity ratio = 4
effort = 40 kgf
distance through which load is raised =15
Velocity ratio
$V . R = \frac{d i s \tan c e t h r o u g h e f f o r t i s a p p l i e d}{d i s \tan c e t h r o u g h w h i c h l o a d i s r a i s e d} 4 = \frac{d_{E}}{15} d_{E} = 4 \times 15 = 60 m ∴ D i s \tan c e t h r o u g h w h i c h e f f o r t i s a p p l i e d = 4 \times 15 = 60 m$
Work done by effort
$W o r k d o n e b y e f f o r t = E f f o r t \times E f f o r t a r m = 40 \times 60 = 2400 J$
Hence work done by effort = 2400 J

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Similar questions

A pulley system with a velocity ratio of 4 is used to lift a load of 100kgf through a vertical height is 15m. The effort required to do so is 40kgf which is applied in the downward direction. calculate the Distance through which effort is applied?

A pulley system with $V . R = 4$ is used to lift a load of $175 k g f$ through a vertical height of $15 m$ . The effort required is $50 k g f$ in the downward direction. $(g = 10 N / k g)$ .

Calculate:

(I) Distance moved by the effort.

(ii) Work done by the effort.

(iii) M.A. of the pulley system.

(iv) Efficiency of the pulley system

A pulley system with $VR = 4$ is used to lift a load of $175 kgf$ through a vertical height of $15 m$ . The effort required is $50 kgf$ in the downward direction. Calculate
(i) Distance moved by the effort.
(ii) Work done by the effort
(iii) $MA$ of the pulley system.
(iv) Efficiency of the pulley system.

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