Arescue worker pulls an injured skier lying on a toboggan (with a combined mass of 127 kg) across flat snow at a constant speed. a 2.43 m rope is attached to the toboggan at ground level, and the rescuer holds the rope taut at shoulder level. if the rescuer's shoulders are 1.65 m above the ground, and the tension in the rope is 148 n, what is the coefficient of kinetic friction between the toboggan and the snow?
from the question we are given the following
combined mass (m) = 127 kg
acceleration (a) = 0 (because speed is constant)
length of the rope = 2.43 m
height of the rescuers shoulder = 1.65 m
tension in the rope (T) = 148 N
acceleration due to gravity (g) = 9.8 m/s^2
coefficient of kinetic friction (μ) = ?
firstly from the diagram attached we can see that the rope and the rescuer with the ground for a right angled triangle with the rescuer being the opposite side and the rope being the hypotenuse. We can calculate the angle (θ) between the rope and the ground using the formula
sin θ = opposite / hypotenuse
θ = sin^(-1) x (opposite / hypotenuse ) = sin^(-1) x (1.65 / 2.43 )
θ = 42.8 degrees
we can get the coefficient of kinetic friction by applying the formula below
net force = force by the rescuer - force of kinetic friction (force as a result of kinematic friction) .....equation 1
whereforce by the rescuer = tensional force x cos θ = 148 x cos 42.8 = 100.49 Nnet force = mass x acceleration = 127 x 0 = 0 Nforce of kinetic friction = ( normal force - vertical component of tensional force ) x μ
force of kinetic friction = (( 127 x 9.8 ) - ( 148 x sin 42.8)) x μ
force of kinetic friction = (1244.6 - 100.6) x μ = 1144μ
now substituting all the values in to equation 1 we have
0 =100.49 - 1144μ
1144μ = 100.49
μ = 0.088
god i hope this !
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