- #1
mickeymouseho
- 6
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1. A horizontal spring, of constant 12N/m, is mounted at the edge of a lab bench to shoot marbles at targets on the floor 93.0 cm below. A marble of mass 8.3 x 10^-3 kg is shot from the spring, which is initially compressed a distance of 4.0 cm. How far does the marble travel horizontally before hitting the floor?
2. Relevant equations
E = 0.5kx^2
E = 0.5mv^2
delta d = 0.5(acceleration)(delta time)^2
dx = vx x t
E = 0.5kx^2
= 0.5(12N/m)(0.04m)^2
= 0.0096J
E = 0.5mv^2
0.0096J = 0.5(8.3x10^-3kg)(v)^2
Square root of [0.0096J / (0.5x8.3x10^-3kg)] = v
2.31 m/s = v
delta d = 0.5at^2
t = square root of [ (2 times delta d) / a] *accleration = gravity, 9.8m/s
t = square root of [ (2 x 0.93m) / (9.8m/s^2)]
t = 0.436s
dx = vx x t
= 2.31 m/s x 0.436s
= 1.00716m
I've got 1.00716m as the answer. However, the textbook says the answer is 0.66m. I'm guessing I did something wrong at finding the velocity. I can't seem to figure this problem out. Please help. Thanks in advance!
2. Relevant equations
E = 0.5kx^2
E = 0.5mv^2
delta d = 0.5(acceleration)(delta time)^2
dx = vx x t
The Attempt at a Solution
E = 0.5kx^2
= 0.5(12N/m)(0.04m)^2
= 0.0096J
E = 0.5mv^2
0.0096J = 0.5(8.3x10^-3kg)(v)^2
Square root of [0.0096J / (0.5x8.3x10^-3kg)] = v
2.31 m/s = v
delta d = 0.5at^2
t = square root of [ (2 times delta d) / a] *accleration = gravity, 9.8m/s
t = square root of [ (2 x 0.93m) / (9.8m/s^2)]
t = 0.436s
dx = vx x t
= 2.31 m/s x 0.436s
= 1.00716m
I've got 1.00716m as the answer. However, the textbook says the answer is 0.66m. I'm guessing I did something wrong at finding the velocity. I can't seem to figure this problem out. Please help. Thanks in advance!
Gosh this is embarrasing...
