A crate of mass 9.4 kg is pulled up a rough incline with an initial speed of 1.4

Question

A crate of mass 9.4 kg is pulled up a rough incline with an initial speed of 1.48 m/s. The pulling force is 96 N parallel to the incline, which makes an angle of 20.40 with the horizontal The coefficient of kinetic friction is 0.400, and the crate is pulled 4.96 m. (a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate-incline system owing to friction c) How much work is done by the 96-N force on the crate? (d) What is the change in kinetic energy of the crate? e) What is the speed of the crate after being pulled 4.96 m? m/s Need Help? Read ItMaster Submit Answer Save Progress My Notes Ask Your Te -12 points SerPSE9 8.P.017 1041 A D e.e-. @ Type here to search

Explanation / Answer

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 the gravitational force 'mg' into two components, i.e. mgsin and mgcos.

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

W = w1 + w2

W = mgsin*(-4.96) + mgcos (0)

= (9.4)(9.8)sin(20.4)(-4.96) + 0

= - 159.27 J

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

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

Ff = R

R = mgcos

therefore, Ff = mgcos x 0.4

Ff = (9.4) (9.8) cos(20.4) (0.4)

Ff = 34.54 N

W(f) = (34.54) (-4.96)

= - 171.30 J

c) W(F) = F.S

= (96 ) (4.96)

= 476.16 J

d) Initial kinetic energy = [m(u)^2] / 2

K.E (i) = [9.4(1.48)^2 ] / 2

K.E (i) = 10.29 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

= 96 - (9.4)(9.8)sin(20.4) – 34.54

= 29.35 N

acc= Rf / mass

= 29.35 / 9.4

= 3.12 N

By equation of motion,

v^2 = u^2 + 2aS

v^2 = (1.48)^2 + 2(3.12)(4.96)

v = 5.76 m/s

Hence,

K.E.(f) = [m(v)^2] / 2

= (9.4)(5.76)^2 / 2

= 155.87 J

Change in K.E. = K.E.(f) - K.E.(i)

K.E. = 155.87 – 10.29

K.E = 145.58 J

e) We have already solved it in the previous part, and the answer is v= 5.76 m/s

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