Question

A block of mass 2.5 kg is displaced by 2.82 m on a
horizontal smooth surface by applying a constant force
of 16 N at angle 45° with horizontal. Find :
(i) work done by the applied force.
(ii) work done by the gravitational force.
(iii) work done by the normal reaction of the surface.
(iv) net work done on the block.​

Answer 1

Answer:

(1) 31.90 J

(2) 0 J

(3) 0J

(4) 31.90J

Explanation:

(1)

by definition of work

w=force × displacement into direction of force

so

work done by applied force

=16cos45×2.82

=31.90J

(2)(3)

so

in vertical direction there are zero displacement and mg (gravitation force) and normal reaction force is working in vertical direction so

W=16cos45×0=0J

(4)

there are only component in horizontal direction of applied force is working and work done by other forces are zero so net work done is

W=31.90J