A 0.75 kg block is going at a velocity of 1.5 m/s and moves on a frictionless surface. Then all...

Question:

A 0.75 kg block is going at a velocity of 1.5 m/s and moves on a frictionless surface. Then all of a sudden, the block hits a rough patch. The rough patch is 0.5 m long and has a coefficient of friction of 0.25. Then, it slides onto the frictionless surface again. Later, the mass collides with a very long spring (spring coefficient is 12,000 N/m). Calculate how far the spring is compressed when the block just comes to rest.

Spring Potential Energy :

Spring potential energy is the energy that is stored in the body due to the work done in order to compress the spring from some initial point to a certain point and the spring potential energy of the body is dependent on the spring constant and the displacement of the spring.

Answer and Explanation:

Given

Mass of the block is {eq}m=0.75\ kg {/eq}

The velocity of the block is {eq}v=1.5\ m/s {/eq}

Length of the rough patch is {eq}l=0.5\ m {/eq}

Coefficient of friction is {eq}\mu = 0.25 {/eq}

Spring constant of the spring is {eq}k=12000\ N/m {/eq}

From the conservation of energy the kinetic energy of the body will be converted into the frictional energy and the spring potential energy:

{eq}S.E-F.E=K.E\\ \frac{1}{2}kx^2-\mu (mg)\times 0.5=\frac{1}{2}mv^2\\ \frac{1}{2}(12000)\times x^2-(0.25\times 0.75\times 9.81\times 0.5)=\frac{1}{2}(0.75)\times (1.5)^2\\ 6000x^2-0.919=0.844\\ x^2=2.93\times 10^{-4}\\ x=0.0171\ m {/eq}

Thus, the displacement of the spring is 0.0171 m


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