A 500 g block is released from rest and slides down a frictionless track that begins h = 1.70 m above the horizontal, as shown in Figure P13.56. At the bottom of the track, where the surface is horizontal, the block strikes and sticks to a light spring with a spring constant of 25.0 N/m. Find the maximum distance the spring is compressed.
Do I find x by: x=sqrt(2*Mgh/k)
yes.
Yes, to find the maximum distance (x) the spring is compressed, you can use the equation:
\[x = \sqrt{\frac{2Mgh}{k}}\]
where:
M = mass of the block (0.500 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = vertical height of the track (1.70 m)
k = spring constant (25.0 N/m)
Plugging in the values into the equation, we have:
\[x = \sqrt{\frac{2 \times 0.500 \times 9.8 \times 1.70}{25.0}}\]
Simplifying:
\[x = \sqrt{\frac{16.52}{25.0}}\]
\[x = \sqrt{0.6608}\]
\[x \approx 0.812\] (rounded to three decimal places)
Therefore, the maximum distance the spring is compressed is approximately 0.812 meters.
Yes, you can find the maximum distance the spring is compressed using the formula x = sqrt(2*M*g*h/k), where:
- x represents the maximum distance the spring is compressed
- M is the mass of the block (500 g or 0.5 kg)
- g is the acceleration due to gravity (9.8 m/s^2)
- h is the height of the track (1.70 m)
- k is the spring constant (25.0 N/m)
To solve for x, plug in the given values into the formula:
x = sqrt((2 * 0.5 kg * 9.8 m/s^2 * 1.70 m) / 25.0 N/m)