Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take g constant with a value 10 m/s2. The work done by the (i) gravitational force and the (ii) resistive force of air is
(i) – 10 J (ii) –8.25 J
(i) 1.25 J (ii) –8.25 J
(i) 100 J (ii) 8.75 J
(i) 100 J (ii) 8.75 J
D.
(i) 100 J (ii) 8.75 J
wg+ wa= Kf– Ki
wa= –8.75 J i.e. work done due to air resistance and work done due to gravity = 10 J
The diagrams below show regions of equipotentials
A positive charge is moved from A to B in each diagram.
Maximum work is required to move q in figure (c).
In all the four cases the work done is the same.
Minimum work is required to move q in figure (a).
Minimum work is required to move q in figure (a).
B.
In all the four cases the work done is the same.
Work was done w = qΔV
ΔV is same in all the cases so work is done will be same in all the cases.
The work done in pulling up a block of wood weighing 2 kN for a length of 10 m on a smooth plane inclined at an angle of 15° with the horizontal is
9.82 kJ
89 kJ
4.35 kJ
5.17 kJ
D.
5.17 kJ
Here :- Weight of block;ω = 2kN; Distance d =10m
Angle of inclination on the plane α =150
The block will be pulled up on a smooth plane
Hence,force of resistance due to the inclination
F =ωsinα = 2×103×Sin 150
=2×103×0.2588
=0.5176 kN
Now work done W = force×displacement =0.1576×103×10=5.17kJ