Asked by ky
A ball of mass m 5
1.80 kg is released from
rest at a height h 5
65.0 cm above a light
vertical spring of force
constant k as in Figure
P5.66a. The ball strikes
the top of the spring
and compresses it a
distance d 5 9.00 cm
as in Figure P5.66b.
Neglecting any energy
losses during the collision, find (a) the speed of the
ball just as it touches the spring and (b) the force con-
stant of the spring.
1.80 kg is released from
rest at a height h 5
65.0 cm above a light
vertical spring of force
constant k as in Figure
P5.66a. The ball strikes
the top of the spring
and compresses it a
distance d 5 9.00 cm
as in Figure P5.66b.
Neglecting any energy
losses during the collision, find (a) the speed of the
ball just as it touches the spring and (b) the force con-
stant of the spring.
Answers
Answered by
Henry
M = 1.80 kg.
h = 65 cm. = 0.65 m.
a. V^2 = Vo^2 + 2g*h. = 0 + 19.6*0.65 = 12.74.
V = 3.57 m/s.
h = 65 cm. = 0.65 m.
a. V^2 = Vo^2 + 2g*h. = 0 + 19.6*0.65 = 12.74.
V = 3.57 m/s.
Answered by
Henry
b. k = F/d = M*g/d = (1.8*9.8)/0.59 = 29.9 N/m.
Answered by
katherine
k=(mgh)/(1/2)(x^2)
k=(1.8)(9.8)(0.09+0.65)/(1/2)(0.09^2)
k=3223.1 N/m
k=(1.8)(9.8)(0.09+0.65)/(1/2)(0.09^2)
k=3223.1 N/m
Answered by
Correct Answers
A) v= square root of: ((2)(g)(h)) => square root of (2*9.8*0.65)
= 3.57 m/s
B) k= (mgh)/(1/2)(x^2)
k=(1.8)(9.8)(0.09+0.65)/(1/2)(0.09^2)
k=3223.1 N/m / 1000
= 3.223 kN/m
= 3.57 m/s
B) k= (mgh)/(1/2)(x^2)
k=(1.8)(9.8)(0.09+0.65)/(1/2)(0.09^2)
k=3223.1 N/m / 1000
= 3.223 kN/m
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