A 0.8 kg block is shot horizontally from a spring, as in the example above, and travels 0.545 m up a long a frictionless ramp before coming to rest and sliding back down. If the ramp makes an angle of 45.0° with respect to the horizontal, and the spring originally was compressed by 0.17 m, find the spring constant.
Assume mechanical (potential + kinetic) energy is conserved after the block is released.
(1/2) k x^2 = M g (0.545 sin45)
Solve for the spring compression x.
There needs to be a smooth transition where the ramp changes from horizontal to 45 degrees up. Otherwise, some emergy will be lost there.
To find the spring constant, we need to use the conservation of energy. The potential energy stored in the compressed spring is given by:
PE_spring = (1/2)kx^2
where k is the spring constant and x is the displacement of the spring.
The initial potential energy stored in the compressed spring is equal to the final potential energy of the block at its maximal height on the ramp. The potential energy of the block is given by:
PE_block = m * g * h
where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the block on the ramp.
Since the block is at rest at the maximal height, its kinetic energy is zero. Therefore, the initial potential energy of the spring is equal to the potential energy of the block at its highest point.
Setting the two equations equal:
(1/2)kx^2 = m * g * h
Plugging in the values given:
(1/2)k(0.17^2) = 0.8 * 9.8 * 0.545
Simplifying the equation:
0.0729k = 4.2744
Now, solve for k:
k = 4.2744 / 0.0729
k ≈ 58.62 N/m
Therefore, the spring constant is approximately 58.62 N/m.