Hi,I have the solution for this problem given. However, I don't understand why the Force of Friction is going in the same direction as the displacement by doing positive e work and also why is not considered in the calculations. Here is the problem:A hover board and rider of total mass 80 kg is moving at 1 m/s and encounters a slope of angle 2o. After moving a distance of 10 m up the slope the linear velocity of the hover board is 2 m/s. An electric motor, which is connected to the axle between the two wheels, produces a driving torque acting on the axle. The total moment of inertia of the two wheels, the axle and the motor is 0.02 kgm2. The wheels have a radius of 100 mm. If the wheels, axle and motor rotate through an angle of 100 radians during the motion find the magnitude of driving torque acting on the axle. Worked Example 3: Rotational Energy and WorkyouWIA hover board and rider of total mass 80 kg is moving at 1 m/sand encounters a slope of angle 2°. After moving a distance of10 m up the slope the linear velocity of the hover board is 2 m/s.An electric motor, which is connected to the axle between thetwo wheels, produces a driving torque acting on the axle. Thetotal moment of inertia of the two wheels, the axle and the motoris 0.02 kgm². If the wheels, axle and motor rotate through anangle of 100 radians during the motion find the magnitude ofdriving torque acting on the axle. Radius of wheels = 100mmPictorialtue2₂10motionV₁motion5=100mTTangle of slopeknownsriderradius ofwheels,total mass of riderm2and hoverboard M = 80kgaccel of gravity; g = 9.81 m/5weight of rider + hoverboard, mginitial velocity of hoverboard, V₁ = 1 m/sfinal velocity of hoverboard, V₂=200/5moment of inertia of motor,axle and wheelsslopeKLbW 2'I=0.02 kyV₂hoverboard.wheels}up the slope, 5=10mdistance moved up the slopeRadius of wheels; M = 100mmangular displacement of wheels, f = 100 radionsangle of slope) 0 = 2°height, hUnknownsdriving torque from motor, Tm=?initial angular velocity of wheelsaxle and motor= ?final angular velocity, W₂ = ?final height above initial position h=2Normal contact force acting on wheels, N=?2 Friction force from slope acting on whells; FR=?height h = 9 sinfdatom forPotential Energyالله= 100 sin (20)= 0.349mJ GonceptualGmgAFRImNtvexFreebody DiagramThe weight of the rider and hoverboard mg does negative work and isconservative (PE: mgh)The normal contact force, Nidoes zero work since perpendicular to motion (unconservative)The static friction for FR does zero work since there is no slip and no motionbetween the wheels and the ground (unconservative)The driving torque, Tm does positive work and is unconservative.We assume the wheels of the hoverboard roll without slipping thereforethe velocity Y = Wr. This is the constraint to the motion.Since there are unconservative forces doing work we willuse the Work - Energy Equation.

Question

Hi,I have the solution for this problem given. However, I don't understand why the Force of Friction is going in the same direction as the displacement by doing positive e work and also why is not considered in the calculations. Here is the problem:A hover board and rider of total mass 80 kg is moving at 1 m/s and encounters a slope of angle 2o. After moving a distance of 10 m up the slope the linear velocity of the hover board is 2 m/s. An electric motor, which is connected to the axle between the two wheels, produces a driving torque acting on the axle. The total moment of inertia of the two wheels, the axle and the motor is 0.02 kgm2. The wheels have a radius of 100 mm. If the wheels, axle and motor rotate through an angle of 100 radians during the motion find the magnitude of driving torque acting on the axle. Worked Example 3: Rotational Energy and WorkyouWIA hover board and rider of total mass 80 kg is moving at 1 m/sand encounters a slope of angle 2°. After moving a distance of10 m up the slope the linear velocity of the hover board is 2 m/s.An electric motor, which is connected to the axle between thetwo wheels, produces a driving torque acting on the axle. Thetotal moment of inertia of the two wheels, the axle and the motoris 0.02 kgm². If the wheels, axle and motor rotate through anangle of 100 radians during the motion find the magnitude ofdriving torque acting on the axle. Radius of wheels = 100mmPictorialtue2₂10motionV₁motion5=100mTTangle of slopeknownsriderradius ofwheels,total mass of riderm2and hoverboard > M = 80kgaccel of gravity; g = 9.81 m/5weight of rider + hoverboard, mginitial velocity of hoverboard, V₁ = 1 m/sfinal velocity of hoverboard, V₂=200/5moment of inertia of motor,axle and wheelsslopeKLbW 2'I=0.02 kyV₂hoverboard.wheels}up the slope, 5=10mdistance moved up the slopeRadius of wheels; M = 100mmangular displacement of wheels, f = 100 radionsangle of slope) 0 = 2°height, hUnknownsdriving torque from motor, Tm=?initial angular velocity of wheelsaxle and motor= ?>final angular velocity, W₂ = ?final height above initial position h=2Normal contact force acting on wheels, N=?2 Friction force from slope acting on whells; FR=?height h = 9 sinfdatom forPotential Energyالله= 100 sin (20)= 0.349mJ GonceptualGmgAFRImNtvexFreebody DiagramThe weight of the rider and hoverboard mg does negative work and isconservative (PE: mgh)The normal contact force, Nidoes zero work since perpendicular to motion (unconservative)The static friction for FR does zero work since there is no slip and no motionbetween the wheels and the ground (unconservative)The driving torque, Tm does positive work and is unconservative.We assume the wheels of the hoverboard roll without slipping thereforethe velocity Y = Wr. This is the constraint to the motion.Since there are unconservative forces doing work we willuse the Work - Energy Equation.

Accepted Answer

Given Data:

To Find:
Why the Force of Friction is going in the same direction as the displacement…