A rope exerts an 18-N force while lowering a 20-kg crate down a plane inclined a

Question

A rope exerts an 18-N force while lowering a 20-kg crate down a plane inclined at 24 (the rope is parallel to the plane). A 24-N friction force opposes the motion. The crate starts at rest and moves 15 m down the plane.

a) Determine the final speed of the crate. The system consists of the crate, the surface of the incline and Earth.

b)Determine the change in gravitational energy.

c)Determine the change in kinetic energy.

d)Determine the change in internal energy.

e)Determine the work done by the rope.

Explanation / Answer

a) net force acting,

Fnet = m*g*sin(20) - T - fk

m*a = m*g*sin(20) - T - fk

a = g*sin(20) - (T + fk)/m

= 9.8*sin(20) - (18+24)/20

= 1.25 m/s

Apply, v^2 - u^2 = 2*a*d

v^2 - 0^2 = 2*1.25*15 (here initial speed , u = 0)

v = sqrt(2*1.25*15)

= 6.12 m/s <<<<<<<<<-----------Answer

b) change in gravtional energy = -m*g*d*sin(20)

= -20*9.8*15*sin(20)

= -1005.5 J <<<<<<<<<-----------Answer

c) change in kinetic energy = 0.5*m*(v2^2 - u^2)

= 0.5*20*(6.12^2 - 0^2) (here initial speed , u = 0)

= 375.5 J <<<<<<<<<-----------Answer

d) change in internal energy = -workdone by friction

= -fk*d*cos(180)

= -24*15*(-1)

= 360 J <<<<<<<<<-----------Answer

e) Workdone by the rope = T*d*cos(180)

= 18*15*(-1)

= -270 J <<<<<<<<<-----------Answer

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