Question

a. A $50g$ice cube can slide without friction up and down a ${30}^o$slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring $10cm$. The spring constant is $25N/m$. When the ice cube is released, what total distance will it travel up the slope before reversing direction?

b. The ice cube is replaced by a $50g$plastic cube whose coefficient of kinetic friction is $0.20$. How far will the plastic cube travel up the slope? Use work and energy.

Short answer

(a) The distance $50g$ice cube will travel up the slope before reversing direction is $51.02cm$.

(b) The distance $50g$public cube will travel up the slope before reversing direction is $37.89cm$when kinetic friction is acting.

Step-by-step solution

Given information (part a)

A $50g$ice cube can slide without friction up and down a ${30}^o$slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring $10cm$. The spring constant is $25N/m$.

Explanation (part a)

By the conservation of energy, the kinetic energy of the ice cube is used up in the gain of K.E. of both the blocks and in doing work against friction.

$\frac{1}{2}m{v}_i^2+mg{h}_i+\frac{1}{2}k{x}_i^2=\frac{1}{2}m{v}_f^2+mg{h}_f+\frac{1}{2}k{x}_f^2$

Since the initial kinetic energy, potential energy, final velocity $v_f$, and spring potential are zero.

$\frac{1}{2}k{x}^2=mgh$

$$\begin{gathered} \text{Find}hbyu\sin gtrigonometry \\ \sin \left(30°\right)=\frac{h}{t} \\ \frac{1}{2}k{x}^2=mgt\sin \left(30°\right) \\ t=\frac{k{x}^2}{2mg\sin \left(30°\right)} \\ t=\frac{\left(25N/m\right)\times {\left(0.1m\right)}^2}{2\times \left(0.05kg\right)\times \left(9.8m/se{c}^2\right)\times 0.5} \\ t=51.02cm \end{gathered}$$

The ice cube will travel up to$51.02cm$

Given information (part b)

A $50g$ice cube can slide without friction up and down a ${30}^o$slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring $10cm$. The spring constant is $25N/m$. The ice cube is replaced by a $50g$plastic cube whose coefficient of kinetic friction is $0.20$.

Explanation (part b)

The energy is partly used up in doing work against gravity. From the work-energy theorem:

$$\begin{gathered} W=-\left(U_f-U_i\right) \\ h=t\left(\sin 30°\right) \\ {U}_f=mgh \\ =mgt\left(\sin 30°\right) \\ {U}_i=\frac{1}{2}k{x}^2 \\ W={\mu }_{k}mgt\cos 30° \\ {\mu }_{k}mgt\cos 30°=-\left(mgt\sin 30°-\frac{1}{2}k{x}^2\right) \\ \frac{1}{2}k{x}^2=mgt\sin 30°+{\mu }_{k}mgt\cos 30° \\ t=\frac{\frac{1}{2}k{x}^2}{mg\sin 30°+{\mu }_{k}mg\cos 30°} \\ t=\frac{0.5\times \left(25N/m\right)\times {\left(0.1m\right)}^2}{\left(0.05kg\right)\times \left(9.8m/se{c}^2\right)\times 0.5+\left(0.05kg\right)\times \left(9.8m/se{c}^2\right)\times 0.2\times 0.87} \\ t=37.89cm \end{gathered}$$

The plastic cube will travel up to $37.89cm$