Question

A $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{kg}$
block is set into motion up an inclined plane with an initial speed of ${\mathbf{v}}_{\mathbf{i}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$(Fig. P8.23).The block comes to rest after traveling$\mathbf{d}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$along the plane, which is inclined at an angle of $\mathbf{\theta }\mathbf{=}\mathbf{30}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ }}$to the horizontal. For this motion, determine (a) the change in the block’s kinetic energy, (b) the change in the potential energy of the block Earth system, and (c) the friction force exerted on the block (assumed to be constant). (d) What is the coefficient of kinetic friction?

Short answer

(a) The kinetic energy change in the block is $∆K=-160\mathrm{J}$.

Step-by-step solution

Step 1:

When two bodies in contact, are in relative motion, a force develops on the surface of contact which tends to oppose the relative motion. This force is called frictional force. It always acts in the direction, opposite to that of relative motion. The work done by friction is stored in the form of internal energy in the body.

$\mathbf{-}\mathbf{∆}{\mathbf{E}}_{\mathbf{int}}\mathbf{=}\mathbf{∆}{\mathbf{E}}_{\mathbf{mech}}\mathbf{=}\mathbf{∆}\mathbf{K}\mathbf{+}\mathbf{∆}{\mathbf{U}}_{\mathbf{g}}\mathbf{=}\mathbf{-}{\mathbf{f}}_{\mathbf{k}}\mathbf{d}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{16}\mathbf{)}$

in the presence of non-conservative forces, the law of conservation of energy is given as-

$\mathbf{\sum }{\mathbf{W}}_{\mathbf{other}\mathbf{}\mathbf{forces}}\mathbf{=}\mathbf{∆}\mathbf{K}\mathbf{+}\mathbf{∆}\mathbf{U}\mathbf{+}\mathbf{∆}{\mathbf{E}}_{\mathbf{int}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(8.17\right)$

where,

$∆\mathrm{k}=$Change in kinetic energy

$∆\mathrm{U}=$Change in potential energy

$∆{\mathrm{E}}_{\mathrm{int}}=$Change in internal energy

Given Data

The mass of the block is- 5 kg

The initial speed is- 8 m/s

Distance covered before coming to rest is- 3 m

Inclination is-$30.0^{\circ }$

Part-(a)

The change in kinetic energy between the initial and final points, is given as-

$$\begin{gathered} ∆K=k_f-k_i \\ =\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2 \\ =0-\frac{1}{2}\left(5.00\mathrm{kg}\right){\left(8.00\mathrm{m}/\mathrm{s}\right)}^2 \\ =-160\mathrm{J} \end{gathered}$$

The change in kinetic energy between initial and final point is -160 J.