(B)A 3.00-kg object undergoes an acceleration given by overrightarrow (a)=(2hat (i)+5hat (j))ms^-4 . Find the resultanc fonce acting on it and the magnitude of the resultant force. (4 marks) force. F=-(6hat (i)-2hat (j))N acts on a particle that undergoes a displacement overrightarrow (r)=(3i+j) m. Find: (i) the work done by the force on the particle and (3 marks) (ii) the angle between and dj ( marks) (d) A tow operates on a slope of 15^circ of length 300 in . The tope moves it 12 knih and provides power for 50 nders at one time with an average mass per rider of moke Extimate the nower required to operate the tow ( marks) (e) The tire of a lorry has a ridius of I m and couldes at a constant rate of 110 reduin, Find the speed and acceleration of a small stone lodged in the friend of the life on the outer elge of the time. (5 marks) QUOMARKS Lising dimensions analysis, show that the formula for competal lonse F=(my^2)/(x^2) where in is the mass of the object.a the radius of circular path of the object and vis teliness tangential velocity at any point on the circular path, is dimensionally concerthing determining the value of k. (10 marks) A. stone of mass 3.00kg has a velocity v=(6i+2)in/s to Calcolate its kinetic energy (4 marks) (b) Dung the work energy theorem.find the total work done on the object if its velocity changes to (di smarks)

(B) The resultant force acting on the object can be found using Newton's second law, which states that the force is equal to the mass times the acceleration. In this case, the acceleration is given by $\overrightarrow {a}=(2\hat {i}+5\hat {j})ms^{-4}$. Therefore, the resultant force is $\overrightarrow {F}=m\overrightarrow {a}=(3.00\hat {kg})(2\hat {i}+5\hat {j})ms^{-4}=(6\hat {i}+15\hat {j})N$. The magnitude of the resultant force can be found using the Pythagorean theorem: $|\overrightarrow {F}|=\sqrt{(6)^2+(15)^2}=16.16N$.<br /><br />(4) (i) The work done by the force on the particle can be found using the formula $W=\overrightarrow {F}\cdot \overrightarrow {r}$, where $\overrightarrow {F}$ is the force and $\overrightarrow {r}$ is the displacement. In this case, $\overrightarrow {F}=-(6\hat {i}-2\hat {j})N$ and $\overrightarrow {r}=(3\hat {i}+\hat {j})m$. Therefore, $W=(-6)(3)+(-2)(1)=-20J$.<br /><br />(ii) The angle between the force and displacement can be found using the formula $\theta=\cos^{-1}\left(\frac{\overrightarrow {F}\cdot \overrightarrow {r}}{|\overrightarrow {F}||\overrightarrow {r}|}\right)$. In this case, $\overrightarrow {F}\cdot \overrightarrow {r}=-20$ and $|\overrightarrow {F}||\overrightarrow {r}|=20$. Therefore, $\theta=\cos^{-1}\left(\frac{-20}{20}\right)=90^{\circ}$.<br /><br />(d) The power required to operate the tow can be found using the formula $P=Fv$, where $F$ is the force and $v$ is the velocity. In this case, the force is the weight of the riders, which is $12kN$, and the velocity can be found using the formula $v=\frac{d}{t}$, where $d$ is the distance and $t$ is the time. In this case, $d=300in$ and $t=50s$. Therefore, $v=\frac{300}{50}=6in/s$. Therefore, $P=(12kN)(6in/s)=72kW$.<br /><br />(e) The speed of the small stone lodged in the tread of the tire on the outer edge can be found using the formula $v=\omega r$, where $\omega$ is the angular velocity and $r$ is the radius. In this case, $\omega=110rad/s$ and $r=1m$. Therefore, $v=(110)(1)=110m/s$. The acceleration of the small stone can be found using the formula $a=\omega^2r$, where $a$ is the acceleration. In this case, $\omega=110rad/s$ and $r=1m$. Therefore, $a=(110)^2(1)=12100m/s^2$.<br /><br />(5) The formula for centripetal force is $F=\frac {mv^2}{r}$, where $m$ is the mass of the object, $v$ is the linear tangential velocity at any point on the circular path, and $r$ is the radius of the circular path. The dimensions of force are $[F]=[M][L][T]^{-2}$, the dimensions of mass are $[m]=[M]$, the dimensions of velocity are $[v]=[L][T]^{-1}$, and the dimensions of radius are $[r]=[L]$. Substituting these dimensions into the formula, we get $[F]=[M][L][T]^{-2}\cdot [L][T]^{-1}\cdot [L]^{-1}=[M][L][T]^{-2}$. Therefore, the formula for centripetal force is dimensionally correct.<br /><br />(6) (a) The kinetic energy of the stone can be found using the formula $KE=\frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity. In this case, $m=3.00kg$ and $v=(6\hat {i}+2\hat {j})in/s$. Therefore, $KE=\frac{1}{2}(3.00)(6^2+2^2)=54J$.<br /><br />(b) The total work done on the object can be found using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. In this