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Work on an incline plane. A 6.0 kg block is pushed 8.0 m up a rough 37 degrees inclined plane by a horizontal force of 75 N. If the initial speed...
Work on an incline plane. A 6.0 kg block is pushed 8.0 m up a rough 37 degrees inclined plane by a horizontal force of 75 N. If the initial speed of the block is 2 m/s up the plane and a constant kinetic friction force of 25 N opposes the motion, calculate: a.) The initial kinetic energy of the block b.) The work done by the 75N force c.) The work done by the friction force d.) The work done by gravity e.) The work done by the normal force f.) What is the net work done on the block
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Student Answers
SAMISOOP | STUDENT
mrs. college teacher didnt see that the mass is 6.0 kg not .6 kg huh?
NEELA | STUDENT
a)The initial kinetic enrgy of the block is the product of Exerted force along the movent of the block and the velocity of the block (1/2)mv^2 = (1/2) 6*2^2 =12 J.
b)The 75 N force makes block move along the inclination for 8m inclined to the direction of force. So the work done is the product of the component of 75N along the inclination and the distance of 8meter = (75Ncos37)*8 = 479.1813 J
c)
Work done by the frictional force = (-25N)(-8m) = 200 J . The negative force worked against the direction of force, It is work done against the movement.
d)
The work done by the gravity: The gravitational force = 6kg(-g)* along 8sin37 but against the direction of gravitational force = 283.3826J.
d) The nomal force: {(6kg *cos37)+75N sin 37} * cos 90 = 0
e)
The net work done by the conservation of energy PE + KE = constant. Therefore after pushing 8 meter incline, the net energy spent is the gained potential energy by 6kg mass from ground level to a height of (8sin37) meter=mgh = 6kg*g*8sin 37 = 283.3827J
KRISHNA-AGRAWALA | STUDENT
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