A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.4

Question

A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.44 m/s. The pulling force is 106 N parallel to the incline, which makes an angle of 20.2 degree with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.02 m. How much work is done by the gravitational force on the crate? J Determine the increase in internal energy of the crate-incline system owing to friction. J How much work is done by the 106-N force on the crate? J What is the change in kinetic energy of the crate? J What is the speed of the crate after being pulled 5.02 m? m/s

Explanation / Answer

First of all, draw the free body-diagram -

Then calculate -

(a) W = mgsin*(-5.35) + mgcos (0)
= (9.2)(9.8)sin(20.2)(-5.02) + 0
= - 156.28 J
So, the work done by gravitational force = -156.28 J

(b) The work done by frictional force(Ff) is the mechanical energy lost..

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

Ff = R

Again -
R = mgcos

therefore, Ff = mgcos x 0.4
Ff = (9.2) (9.8) cos(20.2) (0.4)

Ff = 33.84 N

W(f) = (33.84) (-5.02)
= - 169.88 J

So, the energy lost due to friction = 169.88 J

(c) W(F) = F.S
= (106 ) (5.02)
= 532.12 J

Therefore, the work done by the Force = 532.12 J

(d) Initial kinetic energy = [m(u)^2] / 2
K.E (i) = [9.2*(1.44)^2 ] / 2
K.E (i) = 9.54 J

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)~
Rf = F - mgsin - mgcos
= 106 - (9.2)(9.8)sin(20.2) - 33.84
= 106 - 31.13 - 33.84
= 41.03 N

acc= Rf / mass
= 41.03 / 9.02
= 4.55 m/s^2

By equation of motion,

v^2 = u^2 + 2aS
v^2 = (1.44)^2 + 2(4.55)(5.02)
v^2 = 2.07 + 45.68
v^2 = 47.75

v = 6.91 m/s

Hence,
K.E.(f) = [m(v)^2] / 2
= (9.02)(6.91)^2 / 2
= 215.36 J

Change in K.E. = K.E.(f) - K.E.(i)
K.E. = 215.36 - 9.54
K.E = 205.82 J

Ans. the change in K.E. = 205.82 J

(e) Already solved it in the previous part, and the answer is v= 6.91 m/s

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