ICSE Class 10 Physics 2017 Paper

A pulley system with $V R=4$ is used to lift a load of $175\,\text{kgf}$ through a vertical height of $15\,\text{m}$ . The effort required is $50\,\text{kgf}$ in the downward direction. $(g=10\,\text{N}\,\text{kg}^{-1})$
Calculate:

Question : 28 of 65

Marks: +1, -0

Distance moved by the effort.

Solution:

Given : V.R.

=4

, Load

=175\ \mathrm{kgf}

, displacement of load

(d_{L})=15\ \mathrm{m}

., Effort

=50

kgf,

g=10\ \mathrm{Nkg}^{-1}

Distance moved by the effort

(d_{E})=

We know,

\;\text{ V.R. }\;\;=\;\frac{d_{E}}{d_{L}}

\Rightarrow\;\;d_{E}\;=\;\text{ V.R. }\;\times d_{L}=4\times 15

\;=60\ \mathrm{m}

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