A child of mass m starts from rest and slides without friction from a height h a

Question

A child of mass m starts from rest and slides without friction from a height h along a curved waterslide (see figure). She is launched from a height h/5 into the pool.

(a) Is mechanical energy conserved?

Yes/No    

Why?


(b) Give the gravitational potential energy associated with the child and her kinetic energy in terms of mgh at the following positions: the top of the waterslide, the launching point, and the point where she lands in the pool.

(c) Determine her initial speed v0 at the launch point in terms of g and h.

(d) Determine her maximum airborne height ymax in terms of h, g, and the horizontal speed at that height, vx.

(e) Use the x-component of the answer to part (c) to eliminate vx from the answer to part (d), giving the height ymax in terms of g, h, and the launch angle .

(f) Would your answers be the same if the waterslide were not frictionless?

Yes/No    

Explain.

potential energy kinetic energy top of the waterslide launching point in the pool

Explanation / Answer

a)

mechanical energy is conserved.

because there is no external force acting on the system

hence work done on the system=0

b)at top of the waterslide, height=h

hence potential energy=m*g*h

speed =0

hence kinetic energy=0

total energy=m*g*h

at launching point, height=h/5

hence potential energy=m*g*h/5

speed =v (let)

hence kinetic energy=0.5*m*v^2

as total energy remains constant,

(m*g*h/5)+0.5*m*v^2=m*g*h

==>kinetic energy=4*m*g*h/5

in the pool, height =0

hence potential energy=0

let speed be v1.

then kinetic energy=total energy-potential energy=0.5*m*v1^2=m*g*h

c)

as obtained earlier in part b,

at launching point,

kinetic eenrgy=0.5*m*v0^2=4*m*g*h/5

==>v0=sqrt(8*g*h/5)

d)

horizontal speed=vx=v0*cos(theta)

at maximum height be d.

then at point vertical component of veloicty is 0.

horizontal component of velocity=vx

then potential energy+kinetic energy=total energy

==>m*g*d+0.5*m*vx^2=m*g*h

==>d=(g*h-0.5*vx^2)/g

hence ymax=(g*h-0.5*vx^2)/g

e)

using vx=v0*cos(theta)=sqrt(8*g*h/5)*cos(theta)

we get

ymax=(g*h-0.5*1.6*g*h*cos^2(theta))/g

=> ymax=(g*h-0.8*g*h*cos^2(theta))/g

f)
no answer would not remain the same

because energy have to be spent in order to overcome the friction

hence total mechanical energy of the system will reduce.

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