Is someone able to help me with the following:

Q2 a Track Radius of truck 42.M - centrifugal force acting on Direction of the go kart can be worked out frictional gone using F = mv2 / r2) 6 ) Frax = 0. 2 x (450 kg + 70 kg) x 9. 81 M /5 2 = 0. 2 x 5 20 kg x 9. 81m 15 2 = 1020. 24 N Using F = IN F = frictional force Where N= ( 450 k g + 70 kg ) x 9. 81 m /52 * = coefficient of friction = 5101. 2 N N = Normal force IC F =IN Then Y = = :. 1020 , 24 N 5101 . 20 N = 0.22 ) C ) Rearrange F = M V2 /r M = total mass to V = V (Frim ) V = velocity ( speed ) r = radius Then V = V 1020. 24 N X 42m / 520 kg F = frictional force = 9. 07766 M/S Now rearrange v = r W to w = - Then W = 4.07766 MIS 4 2 m = 0. 2161 rad/s ( to 4d. P. ) Your go karts have a mass of 450kg and travel around a circular curve on a at, horizontal track at a radius of 42 m. a] Draw a diagram to show the go kart on the track and add an arrow to show the direction of the frictional force needed for the car to travel around the curve at a radius of 42 m. h] The maximum frictional force between the tyres and the road is equal to 20% of the weight of the car and driver. Calculate the coefcient of friction when an adult of mass Tkg is driving the kart. c} Calculate maximum angularvelocitv at which the car can travel round the curve at a constant radius of 42 m. d] Calculate the maximum linear velocity. e} You decide that this is not nearlyr fast enough and decide to create a banked track of 120. what is the maximum linear velocity now"? i} Describe qualitatively how this could change if a child were driving the kart. g} It is important that the karts can brake effectively. They use drum brakes where a solid circular drum of mass 4J2} kg and radius 1.15 m is rotating at an angular speed of 22 rad s'1 about an axis when a 'braking' torque is applied to it which brings itto rest in 5.3 s. Calculate it its angular deceleration when the braking torque is applied. ii] the moment of inertia of the drum about the axis shown : lmrE 2 iii} the resultant torque that causes it to decelerate