A block on a frictionless slope slides down and compresses a spring

In summary, the conversation discusses finding the displacement of a spring when a block is projected onto it on an inclined plane. The solution involves calculating the block's kinetic energy at the point of impact, as well as the potential energy of the spring. However, the calculated answer is incorrect and the conversation also suggests using Hooke's law to determine the correct displacement.
  • #1

Homework Statement

Hello, folks. I know this same problem has been answered before here but the previous responses are just not doing it for me, sorry!

An inclined plane of angle θ = 20.0° has a spring of force constant k = 495 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m = 2.37 kg is placed on the plane at a distance d = 0.318 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?

Here is an image of such a setup.

2. The attempt at a solution

So basically I want to find the displacement in the spring given these values. What I do first is get the velocity at which the box hits the spring so I can get it's kinetic energy at that point.

It starts at .750m/s but it's acceleration due to gravity is 9.8*sin(20 degrees) since it's on an incline and it has to travel .318m to hit the spring so I've got:


.318=0+0.75*t+1/2*(9.8*sin(20 degrees))*t^2

t=0.26951 seconds

now that we have time let's get final velocity.


v_final=0.75+9.8*sin(20 degrees)*0.26951

v_final=1.653 m/s

Now let's get the kinetic energy of the block when it hits the spring.



Now let's get the potential energy on the spring



495/2*x^2 J

now let's solve


x=0.114m. This is incorrect.

Where have I gone wrong?

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  • #2
Could you perhaps try and use hookes law to see how much it compresses directly? F=-kx, x=(2.37*9.8*sin(20))/(495)=.016? its been a while since I did physiscs so I am not 100% sure..

is the real answer really close to what you have? it could be a rounding error..

1. How does the slope affect the motion of the block and the compression of the spring?

The slope plays a crucial role in determining the acceleration of the block as it slides down. A steeper slope will result in a higher acceleration, thus causing the block to compress the spring with more force. On the other hand, a gentler slope will lead to a slower acceleration and less compression of the spring.

2. What factors impact the amount of compression in the spring?

The amount of compression in the spring is influenced by a few key factors. These include the mass of the block, the steepness of the slope, and the spring constant which measures the stiffness of the spring. Additionally, the initial velocity of the block and the distance it travels down the slope will also affect the amount of compression.

3. How does the frictionless surface affect the motion of the block and the compression of the spring?

The absence of friction on the slope allows the block to slide down smoothly and with minimal resistance. This results in a consistent acceleration and a predictable amount of compression in the spring. If friction were present, it would slow down the block's motion and decrease the amount of compression in the spring.

4. Can the compression of the spring be used to calculate the mass of the block?

Yes, the compression of the spring can be used to determine the mass of the block using the equation F = kx, where F is the force applied to the spring, k is the spring constant, and x is the distance the spring is compressed. By measuring the force and the compression, the mass of the block can be calculated.

5. How does the energy of the system change as the block slides down and compresses the spring?

As the block slides down the slope, its potential energy decreases while its kinetic energy increases. When the block compresses the spring, its kinetic energy is converted into potential energy in the spring. Therefore, the total energy of the system remains constant, but it is transferred between potential and kinetic energy forms throughout the motion.

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