16. A block of mass 2:0 kg kept at rest on an inclined plane
of inclination 37° is pulled up the plane by applying a
constant force of 20 N parallel to the incline. The force
acts for one second. (a) Show that the work done by the
applied force does not exceed 40 J. (b) Find the work
done by the force of gravity in that one second if the
work done by the applied force is 40 J. (c) Find the
kinetic energy of the block at the instant the force ceases
to act. Take g = 10 m/s?.
Answers
Answer:
Given in the question :-
Force F = 20N .
Mass m = 2.0 kg
Initial velocity u = 0
acceleration, a = 10m/s²
t= 1 sec.
Force works on the block :-
Weight, W = mg
W = 2 × 10
W = 20N (which is Downward)
Normal force N = mg cos37°
N = 20 × 0.80
N = 16 N. (perpendicular & upward to the plane )
Here Applied Force , P = 20N (which is down along the plane)
Now For Final Speed, We know the formula :-
v = u + at
v = 0 + 10× 1
v = 10 m/s
the Distance travelled s = ut +
s= 0 + 0.5 × 10 × 1 × 1
s = 5 m.
Now,
(a) So work done by the force of gravity in 1 sec. = F × d
= 20 N × 5m
= 100 J.
(b) Here the weight act as downward , so distance travelled in downward.
= 5 × sin37°
= 5 × 0.6
= 3 m.
so work done by gravity,
= 20 N × 3 m
= 60 J.
(c) Now, work done by all the forces = change in Kinetic energy
= 1/2 m( v²-u²)
= 0.5 × 2.0 × ( 10² -0²)
= 100 J.
W.D by frictional force = work done by all forces -( work done by Normal force + work done by applied force + work done by gravity )
= 100 J - (100 + 60 +0 )
= 100 - 160
= -60 J
Explanation:
Hope it Helps.