Problem 7.42 A slingshot obeying Hooke\'s law is used to launch pebbles vertical
ID: 1436471 • Letter: P
Question
Problem 7.42 A slingshot obeying Hooke's law is used to launch pebbles vertically into the air. You observe that if you pull a pebble back 14.8 cm against the elastic band, the pebble goes 6.3 m high Part A Assuming that air drag is negligible, how high will the pebble go if you pull it back 29.6 cm instead? Express your answer to two significant figures and include the appropriate units. h= 1 Value Units Submit Give Up Part B How far must you pull it back so it willreach 12.6 mi Express your answer to three significant figures and include the appropriate units.Explanation / Answer
Energy of the sling shot is converted to potential energy of the pebbles.
1/2 kA2 = mgh
k is the spring constant, A is the distance to which it is pulled, and h is the height.
k = 2mgh/A2., A = 0.148 m, h = 6.3 m
a)
1/2 kx2 = mgH
x = 0.296 m
H = [1/2 kx2] / mg
Substituting for k,
H = [1/2 (2mgh/A2) x2] / mg
= (x/A)2h
= (0.296/0.148)2 x 6.3
= 25.2 m
b)
H = 12.6 m = 2h
1/2 kx2 = mgH
x2 = 2mgH/k
Substituting for k,
x2 = 2mgH/(2mgh/A2)
= (H/h)A2 = 2A2
x = A x sqrt(2)
= 14.8 x 1.414 = 20.93 cm.
c)
1/2 kA2 = mgh
k and A is same as in the first case.
LHS is a constant
mgh is also a constant, take H as the new height
mgh = 2mHg
H = h/2 = 6.3/2
= 3.15 m