First slide

Work

Question

A cord is used to lower vertically a block of mass M by a distance d with constant downward acceleration $\frac{g}{4}$ . Work done by the cord on the block is

Moderate

A

$Mg \frac{d}{4}$
B

$3 Mg \frac{d}{4}$
C

$- 3 Mg \frac{d}{4}$
D

Mgd

Solution

When the block moves vertically downward with acceleration $\frac{g}{4}$ then tension in the cord
$T = M (g - \frac{g}{4}) = \frac{3}{4} Mg$
Work done by the cord = $\vec{F} . \vec{s} = Fscosθ$
$= Tdcos (180 °) = - (\frac{3 Mg}{4}) \times d = - \frac{3}{4} Mgd$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App