You must show all work to get full credit 1) Th e coefficient of kinetic frictio

Question

You must show all work to get full credit 1) Th e coefficient of kinetic friction between the inclined plane and the block is 0.7. The s a mass of 15 Kg. It is observed that when the block passes a point located 2 a) What is the normal force of the block? b) What is the friction force acting on the block? m up along the plane, it is moving at a velocity of 1.6 m/seo. c) What is the acceleration of the block? d) What is the velocity of the block as it passes a point 5 m up on the plane? 135 N 2 35% 25° 2) If a bicycle rider is supposed to maintain contact a loop-the-loop ride and the path of the ride is a circle with a radius of 2.7 m. a) What is the minimum velocity that the rider has to keep to maintain contact? 3) Masses m1 (2 Kg) and m2 (7 Kg) are connected by a 13 m long cord, hung over a perfect pulley as shown. The coefficient of kinetic friction between m1 and the plane is 0.4. Mass m1 is released from rest when m2 is located 6 m below the pulley. Note: The pulley does not play a role a) What is the normal force acting on the system? b) What is the friction force acting on the system? c) What is the Tension in the cord? d) What is the acceleration of the system? e) How long does it take for m1 to hit the pulley? f) What is the velocity of m1 as it hits the pulley?

Explanation / Answer

Question 1

a) Along the y direction, the forces acting are Normal reaction (N, upwards), 15cos25(downwards, due to the weight of the block), 135 sin35 upwards.. But since there is no acceleration along the y direction, all the forces add up to zero

N - 15cos25 + 135 sin35 = 0 therefore, N = 15 cos25 - 135 sin 35 = 63.83 N (upward direction)

Force of friction = coeff of friction * N = 0.7*63.83 = 44.681 N

Along the incline the forces produce non zero acceleration

135cos(35) - 15sin(25) - ( 0.7*63.83) = 15a (assuming acceleration along the incline is a )

a = 3.971 m/s2

At 2 mts up the plane the velocity is 1.6 m/s , therefore this is the intitial velocity, for more 3 mts to reach 5mts in all
the final velocity needs to be calculated

using kinematic equations of motion, 2as = v2 - u2

2*3.971*3 = v2 - 1.62

v = 5.13 m/s

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