The block moves at constant speed, so the pulling force is equal to the frictional force.

The normal reaction is

\[ R=200g=200(10)=2000\text{ N}. \]

The frictional force is

\[ F=\mu R=0.8(2000)=1600\text{ N}. \]

Power is given by

\[ P=Fv. \]

So

\[ 4000=1600v. \]

Therefore

\[ v=2.5. \]

Answer: \(2.5\text{ m s}^{-1}\).

(a) For the first \(10\text{ s}\),

\[ u=0,\quad a=0.5,\quad t=10. \]

So the speed after \(10\text{ s}\) is

\[ v_1=u+at=0+0.5(10)=5\text{ m s}^{-1}. \]

For the next part of the motion,

\[ u=5,\quad a=1.5,\quad s=13. \]

Using \(v^2=u^2+2as\),

\[ v^2=5^2+2(1.5)(13)=25+39=64. \]

Therefore

\[ v=8. \]

Answer: \(v=8\text{ m s}^{-1}\).

(b)(i) The distance travelled in the first \(10\text{ s}\) is

\[ s_1=\frac12(0.5)(10)^2=25\text{ m}. \]

The next distance is \(13\text{ m}\).

The distance travelled at constant speed is

\[ s_3=8(5)=40\text{ m}. \]

The final distance is \(28\text{ m}\).

Hence the total distance is

\[ 25+13+40+28=106. \]

Answer: \(106\text{ m}\).

(b)(ii) The time for the second part is

\[ t_2=\frac{8-5}{1.5}=2\text{ s}. \]

For the final part, the cyclist decelerates from \(8\text{ m s}^{-1}\) to rest over \(28\text{ m}\). The average speed during this part is

\[ \frac{8+0}{2}=4\text{ m s}^{-1}. \]

So the time for the final part is

\[ t_4=\frac{28}{4}=7\text{ s}. \]

Total time:

\[ 10+2+5+7=24\text{ s}. \]

Average speed:

\[ \frac{106}{24}=4.416\ldots \]

Answer: \(4.42\text{ m s}^{-1}\).

Resolve horizontally.

The horizontal component of the \(20\text{ N}\) force is \(20\cos\theta\).

The horizontal component of the \(2F\text{ N}\) force is \(2F\cos30^\circ\).

So

\[ 20\cos\theta=2F\cos30^\circ. \]

Since \(\cos30^\circ=\frac{\sqrt3}{2}\),

\[ 20\cos\theta=F\sqrt3. \tag{1} \]

Resolve vertically.

The upward component of the \(20\text{ N}\) force is \(20\sin\theta\).

The downward forces are \(F\) and \(2F\sin30^\circ\).

So

\[ 20\sin\theta=F+2F\sin30^\circ. \]

Since \(\sin30^\circ=\frac12\),

\[ 20\sin\theta=F+F=2F. \]

Therefore

\[ F=10\sin\theta. \tag{2} \]

Substitute (2) into (1):

\[ 20\cos\theta=10\sqrt3\sin\theta. \]

So

\[ 2\cos\theta=\sqrt3\sin\theta. \]

Hence

\[ \tan\theta=\frac{2}{\sqrt3}. \]

Therefore

\[ \theta=49.1^\circ. \]

Using \(F=10\sin\theta\),

\[ F=10\sin49.1^\circ=7.56. \]

Answer: \(F=7.56\), \(\theta=49.1^\circ\).

The loss in gravitational potential energy is

\[ 1600(10)(250-200)=800000\text{ J}. \]

The increase in kinetic energy is

\[ \frac12(1600)(12^2)-\frac12(1600)(10^2). \]

\[ =115200-80000=35200\text{ J}. \]

So the total work done against the resisting forces is

\[ 800000-35200=764800\text{ J}. \]

The work done against the additional constant force is

\[ 150(2000)=300000\text{ J}. \]

Therefore the work done against the braking force is

\[ 764800-300000=464800\text{ J}. \]

Since the braking force acts opposite to the motion, the work done by the braking force is negative.

Answer: \(-4.65\times10^5\text{ J}\).

(a) Consider the two particles as one system.

The total mass is

\[ 0.3+0.2=0.5\text{ kg}. \]

The component of weight down the plane is

\[ 0.5(10)\sin30^\circ=2.5\text{ N}. \]

The total frictional force is

\[ 0.4(0.5)(10)\cos30^\circ=\sqrt3\text{ N}. \]

Using \(F=ma\) up the plane:

\[ 12-2.5-\sqrt3=0.5a. \]

So

\[ a=15.535\ldots \]

Therefore

\[ a=15.5\text{ m s}^{-2}. \]

Now consider particle \(P\).

For \(P\), the component of weight down the plane is

\[ 0.3(10)\sin30^\circ=1.5\text{ N}. \]

The frictional force on \(P\) is

\[ 0.4(0.3)(10)\cos30^\circ=1.2\cos30^\circ. \]

Using \(F=ma\) up the plane:

\[ T-1.5-1.2\cos30^\circ=0.3a. \]

Substituting \(a=15.535\ldots\),

\[ T=7.20\text{ N}. \]

Answer: acceleration \(=15.5\text{ m s}^{-2}\), tension \(=7.20\text{ N}\).

(b) After the string breaks, \(P\) is still moving up the plane. The forces on \(P\) down the plane are the component of weight and friction.

The deceleration of \(P\) is

\[ 10\sin30^\circ+0.4(10)\cos30^\circ. \]

\[ =5+4\cos30^\circ=8.464\ldots\text{ m s}^{-2}. \]

Using \(v=u+at\), with final velocity \(0\),

\[ 0=2-8.464\ldots t. \]

So

\[ t=0.236\text{ s}. \]

Answer: \(0.236\text{ s}\).

(a)(i) Use conservation of momentum.

Before the collision, the momentum is

\[ 4m(6.5)=26m. \]

After the collision, the momentum is

\[ 4m(2)+3m(v)=8m+3mv. \]

So

\[ 26m=8m+3mv. \]

Divide by \(m\):

\[ 26=8+3v. \]

Therefore

\[ v=6. \]

(a)(ii) Initial kinetic energy:

\[ \frac12(4m)(6.5)^2=84.5m. \]

Final kinetic energy:

\[ \frac12(4m)(2)^2+\frac12(3m)(6)^2=8m+54m=62m. \]

Energy lost:

\[ 84.5m-62m=22.5m. \]

Given that the energy lost is \(45\text{ J}\),

\[ 22.5m=45. \]

So

\[ m=2. \]

Answer: \(m=2\).

(b) After the first collision, \(Q\) moves at \(6\text{ m s}^{-1}\).

The time for \(Q\) to reach the wall is

\[ \frac{9}{6}=1.5\text{ s}. \]

During this time, \(P\) travels

\[ 2(1.5)=3\text{ m}. \]

So when \(Q\) reaches the wall, \(P\) is \(9-3=6\text{ m}\) from the wall.

After rebounding, \(Q\) moves towards \(P\) at \(0.5\text{ m s}^{-1}\), while \(P\) continues towards the wall at \(2\text{ m s}^{-1}\).

The closing speed is

\[ 2+0.5=2.5\text{ m s}^{-1}. \]

The time until the second collision is

\[ \frac{6}{2.5}=2.4\text{ s}. \]

In this time, \(Q\) moves away from the wall by

\[ 0.5(2.4)=1.2\text{ m}. \]

Answer: \(1.2\text{ m}\).

(a) Since

\[ v=t-7t^{\frac12}+12, \]

differentiate to find the acceleration:

\[ a=\frac{dv}{dt}=1-\frac72t^{-\frac12}. \]

At \(t=4\),

\[ a=1-\frac72(4^{-\frac12}). \]

\[ a=1-\frac72\cdot\frac12=1-\frac74=-\frac34. \]

Answer: \(-0.75\text{ m s}^{-2}\).

(b) \(P\) is instantaneously at rest when \(v=0\).

\[ t-7t^{\frac12}+12=0. \]

Let \(u=\sqrt t\). Then \(t=u^2\), so

\[ u^2-7u+12=0. \]

\[ (u-3)(u-4)=0. \]

So

\[ u=3\quad\text{or}\quad u=4. \]

Therefore

\[ t=9\quad\text{or}\quad t=16. \]

Answer: \(t=9\text{ s}\) and \(t=16\text{ s}\).

(c) The velocity changes sign at \(t=9\) and \(t=16\).

First find the displacement function:

\[ s=\int \left(t-7t^{\frac12}+12\right)\,dt. \]

\[ s=\frac12t^2-\frac{14}{3}t^{\frac32}+12t. \]

Evaluate:

\[ s(0)=0, \]

\[ s(9)=\frac{45}{2}, \]

\[ s(16)=\frac{64}{3}, \]

\[ s(25)=\frac{175}{6}. \]

Total distance is

\[ \left(s(9)-s(0)\right)+\left(s(9)-s(16)\right)+\left(s(25)-s(16)\right). \]

\[ =\frac{45}{2}+\left(\frac{45}{2}-\frac{64}{3}\right)+\left(\frac{175}{6}-\frac{64}{3}\right). \]

\[ =\frac{63}{2}=31.5. \]

Answer: \(31.5\text{ m}\).