Since

\[ \sin\theta=\frac5{13}, \]

we have

\[ \cos\theta=\frac{12}{13}. \]

(a) The horizontal component of the tension is

\[ 26\cos\theta=26\left(\frac{12}{13}\right)=24\text{ N}. \]

The vertical component of the tension is

\[ 26\sin\theta=26\left(\frac5{13}\right)=10\text{ N}. \]

Vertically,

\[ R+10=W. \]

So

\[ R=W-10. \]

The frictional force is

\[ 0.25R=0.25(W-10). \]

The mass of the crate is

\[ \frac{W}{10}. \]

Using \(F=ma\) horizontally:

\[ 24-0.25(W-10)=\frac{W}{10}(1.5). \]

\[ 24-0.25W+2.5=0.15W. \]

\[ 26.5=0.4W. \]

\[ W=66.25. \]

Answer: \(W=66.3\text{ N}\).

(b) After the rope is removed, the only horizontal force is friction.

The frictional force is

\[ 0.25W. \]

The mass is \(\frac{W}{10}\), so the deceleration is

\[ a=\frac{0.25W}{W/10}=2.5\text{ m s}^{-2}. \]

Using

\[ v^2=u^2+2as \]

with \(v=0\), \(u=8\), and \(a=-2.5\):

\[ 0=8^2+2(-2.5)s. \]

\[ 0=64-5s. \]

\[ s=12.8. \]

Answer: \(AB=12.8\text{ m}\).

(a) Differentiate \(v\) to find the acceleration:

\[ a=\frac{dv}{dt} =\frac{2k_1}{(4t+1)^{\frac12}}-(2t+1). \]

When \(t=1.25\),

\[ 4t+1=6 \]

and

\[ 2t+1=3.5. \]

The deceleration is \(0.5\text{ m s}^{-2}\), so

\[ a=-0.5. \]

Hence

\[ \frac{2k_1}{\sqrt6}-3.5=-0.5. \]

\[ \frac{2k_1}{\sqrt6}=3. \]

Therefore

\[ k_1=\frac{3\sqrt6}{2}. \]

Answer: \(k_1=\frac{3\sqrt6}{2}\).

(b) Since \(P\) is at instantaneous rest when \(t=3.5\),

\[ v=0 \]

when \(t=3.5\).

So

\[ 0=k_1(15)^{\frac12}-\frac14(8)^2+k_2. \]

Using \(k_1=\frac{3\sqrt6}{2}\),

\[ 0=\frac{3\sqrt6}{2}\sqrt{15}-16+k_2. \]

\[ 0=\frac{9\sqrt{10}}{2}-16+k_2. \]

Therefore

\[ k_2=16-\frac{9\sqrt{10}}{2}. \]

The particle is not at instantaneous rest in \(0\leq t\leq1\), so the distance travelled is

\[ \int_0^1 v\,dt. \]

Thus

\[ \int_0^1\left(k_1(4t+1)^{\frac12}-\frac14(2t+1)^2+k_2\right)\,dt. \]

\[ =\left[\frac{k_1}{6}(4t+1)^{\frac32}-\frac1{24}(2t+1)^3+k_2t\right]_0^1. \]

\[ =\frac{k_1}{6}\left(5^{\frac32}-1\right)-\frac{13}{12}+k_2. \]

Substitute \(k_1=\frac{3\sqrt6}{2}\) and \(k_2=16-\frac{9\sqrt{10}}{2}\):

\[ \text{distance} = \frac{\sqrt6}{4}(5\sqrt5-1)-\frac{13}{12}+16-\frac{9\sqrt{10}}{2}. \]

\[ =6.9205\ldots \]

Answer: \(6.92\text{ m}\).