I am confused as to what to do on this problem. Could someone help me? Thanks!
A 67.8 g ball is dropped from a height of 67.3 cm above a sprin of negligible mass. The ball compresses the spring to a maximum displacement of 4.76386 cm. The acceleration of gravity is 9.8 m/s^2. Calculate the spring force constant k. Answer in units of N/m.
2007-10-10 04:09:41 · 3 answers · asked by sg88 1 in Science & Mathematics ➔ Physics
At the instant of maximum displacement, the ball's gravitational potential energy PE1 = mgh has been converted to spring displacement potential energy PE2 = kx^2/2. The total height of the drop h = 0.673 + 0.0476 m. The displacement x = 0.0476 m. Form the equation expressing the equality of PE1 and PE2 and solve for k.
EDIT: For some reason I get 421.97 N/m doing the exact same calculation as the answer below mine.
2007-10-10 04:34:48 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
Using law of conservation energy we have
!/2 k x.x =mg (h+x )
k = spring constant, m= given mass, h= height of fall
g=accn due to gravity
x= compression of the spring.
now you calculate
2007-10-10 04:36:50 · answer #2 · answered by Apparao V 4 · 0⤊ 0⤋
mgh = .5kx^2
(.0678)(9.8)(.673+.0476386) = .5 k (.0476386)^2
k = 426 N/m^2
2007-10-10 04:36:11 · answer #3 · answered by vcas30 3 · 0⤊ 0⤋