An 8 kg metal block of dimensions 16 cm x 8 cm x 6 cm is lying on a table with its face of largest area touching the table. If g = 10 ms -2 , then the minimum amount of work done in making it stand with its length vertical is
Text Solution
Verified by ExpertsThe correct answer is:
C
When face of the largest area is touching the table, height of center of gravity above the table 
cm
With its length vertical, height of center of gravity would become 
cm
Minimum work required W = (P.E.) 2 - (P.E.) 1
= mg (h 2 – h 1 )
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
A particle is released from a height S. At certain height its kinetic energy is three times its pot…
A vertical spring of force constant 100 Nm -1 is attached to the ceiling, with a block of mass 10 k…
A body is projected at an angle of 60° with the horizontal. If its kinetic energy at its maximum he…
The velocity-time graph of a particle of mass 10 kg is shown in the figure. The net work done on th…
A block of mass m is stationary with respect to the wedge of mass M moving with uniform speed v on …
The work done by the force in displacing an object from to , is