A girl swings on a playgroundswing in such a way that at her highest pointshe is

Question

A girl swings on a playgroundswing in such

a way that at her highest pointshe is 3.1m

from the ground, while at herlowest point she

is 0.5 m from the ground.

What is hermaximum speed? Answer in

Given: In the followingchoices, htop is the

height of the highestpoint and hbot is the

height of the lowestpoint.

At what height above theground will the

girl be moving at a speedhalf of her maximum

speed?

a) h1/2 =(1/2)htop

b) h1/2 =htop

c) h1/2 =(1/4) htop

d) h1/2 = 1/4(3htop + hbot)

e) h1/2 = 1/2(htop + hbot)

f) h1/2 =1/2 (htop)

g) h1/2 = 1/4(htop + hbot)

h) h1/2 =hbot

i) h1/2 =1/2 [htop + (2-1)hbot

j) h1/2 =3/4(htop)

Explanation / Answer

the girl have max speed at the highest point and 0 velocity at thelowest point m=mass of the girl taking potential energy to be 0 at the lowest point potential energy at the highestpoint=m*g*h=m*9.8*(3.1-0.5)=25.48*m v=max velocity kinetic energy at the lowest point=(1/2)*m*v2 according to conservation of energy (1/2)*m*v2=25.48*m v=7.14 m/s let the girl has half the max speed at a height of h m from thelowest point potential energy at height h from lowest point=m*g*h and kineticenergy at that point=(1/2)*m*(v/2)2 m*g*h+(1/2)*m*(v/2)2=(1/2)*m*v2 9.8*h+(1/2)*(7.14/2)2=(1/2)*7.142 h=1.95 m above the lowest point or (1.95+.5)=2.45 m above theground

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