Question

A 2 kg block is held in place in front of a compressed spring. The block is resting on a

frictionless surface with a ramp set at a 30 degree angle. The spring constant is 24 N/m and
the spring was compressed 25 cm. When the spring-mass system is released the mass is
free to move away from the spring. How long does the ramp need to be to keep the block
on the ramp? Assume the block is a point mass and that there is no friction on the ramp.

Answer 1

Using the law of conservation of energy and trigonometry, the ramp needs to be at least 15 cm long to keep the block on the ramp when it's released from the compressed spring with a spring constant of 24 N/m and compressed by 25 cm.

Step-by-step explanation:

Calculating the Ramp Length for a Spring-Mass System:

To determine how long the ramp needs to be to keep the block on the ramp when released from a compressed spring, we need to calculate the potential energy stored in the spring and convert that to kinetic energy as the block moves up the ramp. We can use the conservation of energy for this calculation. The potential energy (PE) stored in the compressed spring is given by PE = ½ kx², where k is the spring constant and x is the compression distance.

Using the given values, k = 24 N/m and x

= 0.25 m (25 cm),

we find that PE = ½ * 24 N/m * (0.25 m)²

= 0.75 J.

As the block moves up the ramp, this energy is converted into gravitational potential energy (GPE), where GPE = mgh. Here, m is the mass of the block, g is the acceleration due to gravity (9.81 m/s²), and h is the height gained along the ramp.

Assuming the angle of the incline is 30°, we use trigonometry to find h in terms of the length of the ramp L: h = L sin(30°). Setting PE equal to GPE and solving for L, we have L = 2PE / (mg sin(30°)).

Using m = 2 kg, we get L = 2 * 0.75 J / (2 kg * 9.81 m/s² * 0.5)

= 0.15 m or 15 cm.

Therefore, the ramp needs to be at least 15 cm long to keep the block on the ramp.