Question

A crate of mass 9.0 kg is pulled up a rough incline with an initial speed of 1.52 m/s. The pulling force is 90 N parallel to the incline, which makes an angle of 20.8

Answer 1

OK, here's the solution,
a) We have the relation, W = F.S
where F is the force and S is the displacement of the body.

We have to find out the work done by the gravitational force, therefore, let us break th gravitational force 'mg' into two components, i.e. mgsin? and mgcos?. (look at the figure)~

http://i726.photobucket.com/albums/ww268...

Since we have broken the gravitational force into two components, therefore~

W = w1 + w2
W = mgsin?*(-5.10) + mgcos? (0)
= (9)(9.8)sin(20.8)(-5.10) + 0
= - 159.7 J

Ans. The work done by gravitational force = -159.7 J
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b) The work done by frictional force(Ff) is the mechanical energy lost..

Ff = Normal reaction ( R ) x Coefficient of friction ( ? )

Ff = R ?

again look at the above figure, and you will find that
R = mgcos?

therefore, Ff = mgcos? x 0.4
Ff = (9) (9.8) cos(20.8) (0.4)

Ff = 32.9 N

W(f) = (32.9) (-5.10)
= - 167.79 J

Ans. The energy lost due to friction = 167.79 J
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c) W(F) = F.S
= (100 ) (5.10)
= 510J

Ans. The work done by the Force = 510 J
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d) Initial kinetic energy = [m(u)^2] / 2
K.E (i) = [9(1.52)^2 ] / 2
K.E (i) = 10.39J

Now, for finding the final kinetic energy of the crate, we need to find out the final velocity of the crate. And for the final velocity, we need to find out the acceleration, and for the acc. we need to find out the resultant force (Rf)~

see the F.B.D.

http://i726.photobucket.com/albums/ww268...

Rf = F - mgsin? - ?mgcos?
= 100 - (9)(9.8)sin(20.8) - 32.9

= 35.7 N

acc= Rf / mass
= 35.7 / 9
= 3.96 N

By equation of motion,

v^2 = u^2 + 2aS
v^2 = (1.52)^2 + 2(3.96)(5.10)
v^2 = 2.31+ 40.39
v^2 = 42.70

v = 6.53m/s

Hence,
K.E.(f) = [m(v)^2] / 2
= (9)(42.7) / 2
= 192.15 J

Change in K.E. = K.E.(f) - K.E.(i)
?K.E. = 192.15 - 10.39
?K.E = 181.76 J

Ans. the change in K.E. = 181.76 J
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e) We have already solved it in the previous part, and the answer is v= 6.53m/s
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hope i helped......