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A 3-kg block sits on an incline where the top half of the incline has a coefficient of kinetic...

Question:

A 3-kg block sits on an incline where the top half of the incline has a coefficient of kinetic friction of 0.5 and the bottom half is frictionless. The angle of inclination is 35 degrees. If the block is released and travels 10 m along the rough part of the incline and then 10 in along the smooth part before it makes contact with the spring (k 200 Nm). Calculate the distance the spring is compressed.

Motion Of The Object On The Inclined Plane:

When an object is placed on the inclined plane then the object starts moving down the plane under the influence of acceleration due to gravity and friction opposes the motion of the object on the inclined plane. The angle of inclination also plays a very important role while deciding the motion of the object on an inclined plane.

Answer and Explanation:

Given

Mass of the block is {eq}m= 3\ kg {/eq}

Coefficient of kinetic friction is {eq}\mu_k= 0.5 {/eq}

The angle of inclination is {eq}\theta = 35^\circ {/eq}

The distance travelled on the inclined plane is {eq}S= 10\ m {/eq}

Spring constant of the spring is {eq}k= 200\ Nm {/eq}

Now for the height of the inclined plane :

{eq}h=10\ sin 35^\circ\\ h=5.735\ m {/eq}

Now for the friction force :

{eq}f=\mu (mg)\cos 35^\circ\\ f=0.5\times (3\times 9.81)\cos 35^\circ\\ f=12.05\ N {/eq}

Now for the energy lost due to friction:

{eq}F.E=f\times 10\\ F.E=12.05\times 10\\ F.E= 120.54\ J {/eq}

For the potential energy of the block at the top of the inclined plane:

{eq}U=mgh\\ U= 3\times 9.81\times 5.73\\ U=168.63\ J {/eq}

According to the conservation of energy, the spring potential energy will be equal to the sum of the potential energy and the frictional energy :

{eq}\frac{1}{2}kx^2=U-F.E\\ \frac{1}{2}(200)(x)^2=168.63- 120.54\\ 100\ x^2=48.09\\ x^2=0.481\\ x=0.693\ m {/eq}

Thus, the distance by which the spring is compressed is 0.693 m


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