How much joule of work shall be
done by a rectangular object of mass
1 kg when it slides down 50 cm on
an inclined plane making an angle of
30° with the horizontal ground ?
(g = 10 m/s2
)
Answers
Answered by
0
Explanation:
Since the particle is sliding down with a constant velocity therefore the net force on the particle is zero. The frictional force (acting up the plane ) is balanced by the component mg sin 30.
f = mg sin 30 = 5x10x0.5 = 25 N
Answered by
0
Answer:
HELLO,
s = 50cm, a = g = 10 m/s², M= 1kg.
Cos Theta = 30°
F = m × a
F = 1 × 10
F = 10 Newton
Now, 50 cm = 50/100 m
Work done by rectangular object = F × s Cos Theta
= 10 × 50 /100 × √(3) /2
= 500/100 × √(3) /2
= 5√(3)/2 Joule
5√(3)/2 Joule of work should be done by rectangular object.
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