Given:
Weight of block: = m = 0.6 kg
Distance moved by block up the inclined plane = 8.0 m
Angle of inclined plane = A = 37 degrees
Force applied on block = F = 75 N
Direction of force applied on block = Horizontal
Initial speed of block = u = 2 m/s
Kinetic friction = k = 25 N
The horizontal force F applied to the block can be resolve in two components: f1, acting along the horizontal, and f2 acting perpendicular direction pressing the block down to the plane.
f1 = F*CosA = 75*Cos37 = 75*0.7986 = 59.895 N
f1 = F*SinA = 75*Sin37 = 75*0.6018 = 45.135 N
Initial kinetic energy of the block
Initial kinetic energy = (m*u^2)/2 = (0.6*2^2)/2 = 1.2 N
Work done by the 75 N force
Work done by the 75 N force = w
= f1*s = 59.895*8 = 479.16 J
(Please note that this work is used in three ways - overcoming friction, increasing velocity of block. and raising block against gravitational force.)
Work done by the friction force
There is no work done by friction force. There is only work done in overcoming friction force. Which is given by formula:
Work done against friction = (Frictional force)*(Distance moved) = 25*8 = 200 J
The work done by gravity
There is no work done by gravity. However work is done against gravity. This is given by:
(Mass of block)*(gravitational force)*(vertical movement of block)
= (Mass of block)*(gravitational force)*(s*Cos 37)
= 0.6*9.81*8*0.6018 = 23.3376 J
The work done by the normal force
There is no movement along a direction perpendicular to the plain. Therefore no work is done by the normal force.
Net work done on the block
Net work done in the block as calculated above is 479.16. This consists of three components:
1) Overcoming friction = 200 J
2) Raising block vertically = 23.3376
3) increasing kinetic energy of block. = 479.16 - 200 - 23.3376 = 255.8224
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