# A block and a solid sphere each of mass 3.0 Kg are moving side by side, each with linear speed of...

## Question:

A block and a solid sphere each of mass 3.0 Kg are moving side by side, each with linear speed of 2.5 m/s on a flat surface. The block is sliding on a frictionless surface, and the sphere has a radius of 0.1 m and is rolling without slipping. After a short while, they encounter a ramp that makes an angle 20 degrees with the horizontal.

A. What are the total kinetic energies of the block and the sphere?

B. What distances will the block and the sphere move up the ramp before coming momentarily to rest? Assume there is negligible friction between the ramp and the block, but that the sphere rolls without slipping up the incline.

C. Assuming that the ramp is replaced with one for which there is negligible friction for both the sphere and the block. How, if at all, does the answer to part B above change?

## Calculating Kinetic Energy For Different Dynamics:

For a body in motion, a body can have linear motion dynamics or a rotational motion dynamics, for a body in linear motion dynamics the body has a kinetic energy of {eq}\frac{1}{2}mv^2 {/eq} and for rotational it will have {eq}\frac{1}{2}I\omega^2 {/eq}, where m is the mass, v is the velocity, I is the moment of inertia and {eq}\omega {/eq} is the angular velocity.

## Answer and Explanation:

(a) The total kinetic energy on the block is

{eq}\displaystyle KE_B=\frac{1}{2}mv^2+\frac{1}{2}mg sin 20 v^2\\ =\frac{1}{2}(3)(2.5)^2+\frac{1}{2}(3)(9.8)sin 20 (2.5)^2\\ =9.375+31.2375\\ =40.6125 \ J {/eq}

The sphere has a condition of no slipping

Therefore we have {eq}\displaystyle v=r\omega\\ \omega=\frac{v}{r}\\ =\frac{2.5}{0.1}\\ =25 \ rad/s {/eq}

The moment of inertia I of the sphere will be {eq}\displaystyle I=mr^2 {/eq}

The total kinetic energy of the sphere is

{eq}\displaystyle KE_s=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2+\frac{1}{2}(mg sin 20)v^2\\ =\frac{1}{2}(3)(2.5)^2+\frac{1}{2}(3\times 0.1^2)(25)^2+\frac{1}{2}(3)(9.8)(0.34)(0.25)^2\\ =\frac{1}{2}((3)(2.5)^2+(3)(0.1)^2(25)^2+3\times 9.8\times 0.34\times 0.25^2)\\ =19.06 \ J {/eq}

(b) Let the distance traveled by each be *x*

So using work-energy theorem for the block

{eq}\displaystyle mg sin 20(x)=\frac{1}{2}v^2(mg sin 20 -m)\\ 3\times 9.8\times 0.34(x)=\frac{1}{2}(2.5)^2(3\times 9.8\times 0.34-3)\\ 9.996(x)=21.8625\\ x=2.1871 \ m {/eq}

Similarly, for the sphere we have

{eq}\displaystyle mg sin 20 (x)=\frac{1}{2}mv^2(g sin 20-(I\omega^2+1))\\ 3\times 9.8\times 0.34(x)=\frac{1}{2}(3)(2.5)^2(9.8\times 0.34-0.03-1)\\ x=2.158 \ m {/eq}

(c) The change in the answer will be there as when the work-energy theorem will be applied the net mass for the final kinetic energy will be that of the block and the sphere.

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Chapter 4 / Lesson 14