A cart of unknown mass is pushed 3.61 cm against a hoop spring (spring constant 59.00 N/m) and released, just as in the lab experiment. It is observed to roll down the track (after losing contact with the spring) at constant speed v.
A second spring has spring constant 80.00 N/m. If the experiment is repeated, by what distance should this second spring be compressed in order for the cart to acquire the same final speed v?
Answer in cm
For the spring, the maximum potential energy is equal to the
maximum kinetic energy
(1/2) k x2 = (1/2) m v2
Where x is the compressed length, v is the final velocity, k is the spring constant and m is the mass of the cart.
The experiment is done with two springs with different spring constants,
In the first case,
(1/2) k1 x12 = (1/2) m v2
In the second case,
(1/2) k2 x22 = (1/2) m v2
In both cases, the final velocity of the cart is the same, We need to find the compression length x2 for the second case.
We can equate the LHS of both the above equations,
(1/2) k1 x12 = (1/2) k2 x22
k1 / k2 = (x2 / x1)2
x2 = x1 sqrt(k1 / k2 )
x2 = 3.61 x sqrt ( 59.00 N/m / 80.00 N/m)
x2 = 3.10 cm
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