Question

A 2 kg block is pushed against a spring (k = 400 N/m), compressing it 0.3 m. When the block is released, it moves along a frictionless, horizontal surface, and then up an incline that has friction (μs = 0.4 and μk = 0.2). How far up the incline does the block slide before coming to rest?

Answer 1

Answer:

$2.29 \mathrm{m} \text { the block slide if the } \mathrm{u}_{\mathrm{s}}=0.4$

$4.58 \mathrm{m} \text { the block slide if the } \mathrm{u}_{\mathrm{k}}=0.2$

Explanation:

Given values  

Mass (m) = 2kg

K = 400 N/M

Compressing it 0.3 m

The law of conservation of energy:

$\frac{m v^{2}}{2}+\frac{k x^{2}}{2}=\text { constant }$

$\text { Where, } \frac{m v^{2}}{2} \text { is kinetic energy of the block. }$

$\frac{k \Delta l^{2}}{2}$ Energy of the spring deformation.

M mass of the block

x spring deformation

Therefore, if block left the spring (x = 0)

$\frac{m v^{2}}{2}+0=0+\frac{k \Delta l^{2}}{2}$

Where, Δl is initial spring deformation

$\frac{m v^{2}}{2}=\frac{k \Delta l^{2}}{2}$

$\mathrm{v}^{2}=\frac{k \Delta l^{2}}{m}$

$\mathrm{v}=\sqrt{\frac{k}{m} \times \Delta l^{2}}$

$v=\sqrt{\frac{400}{2} \times(0.3)^{2}}$

$\mathrm{v}=\sqrt{200 \times 0.09}$

The law of conservation of energy:

$\frac{m v^{2}}{2}+m g h=\text { constant }$

Where h is height

$\frac{m v^{2}}{2}+0=0+m g h$

$\frac{m v^{2}}{2}=m g h$

Cancel mass "m" each side

$\mathrm{h}=\frac{v^{2}}{2 g}$

Distance along incline equals

$\begin{array}{ll}{\text { For friction us }} & {\left(L=\frac{h}{u_{s}}\right)} \\ {\text { For friction } u_{k}} & {\left(L=\frac{h}{u_{k}}\right)}\end{array}$

$\begin{array}{l}{\mathrm{u}_{\mathrm{s}}=0.4} \\ {\mathrm{U}_{\mathrm{k}}=0.2} \\ {\text { For friction } \mathrm{u}_{\mathrm{s}}}\end{array}$

$\begin{array}{l}{\mathrm{h}=\frac{v^{2}}{2 g u_{s}}} \\ {\mathrm{L}=\frac{4.24^{2}}{2 \times 9.8 \times 0.4}} \\ {\mathrm{L}=\frac{17.9776}{784}}\end{array}$

$\begin{array}{l}{L=2.29 \mathrm{m}} \\ {2.29 \mathrm{m} \text { the block slide if the } \mathrm{u}_{5}=0.4} \\ {\text { For friction } \mathrm{u}_{\mathrm{k}}} \\ {\mathrm{L}=\frac{4.24^{2}}{2 \times 9.8 \times 0.2}}\end{array}$

$\begin{array}{l}{L=\frac{17.9776}{3.92}} \\ {L=4.58 \mathrm{m}} \\ {4.58 \mathrm{m} \text { the block slide if the } \mathrm{u}_{5}=0.4}\end{array}$