What minimum coefficient of kinetic friction is required to stop the crate over

Question

What minimum coefficient of kinetic friction is required to stop the crate over a distance of 5.0 m along the lower surface? 0.30 0.32 0.60 0.76 0 66 20. A 4.0 kg Nock and a 2.0 kg Nock can move on a horizontal frictionless surface. The Nocks are accelerated by a +12 N force that pushes the larger Nock against the smaller one. Determine the force that the 2.0 kg Nock exerts on the 40 kg black. zero -4 N + 4 N + 8 N_12 N An automobile approaches a harrier at a speed of 20 m/s The driver locks the brakes at a distance of 50 m from the barrier. What minimum coefficient of kinetic friction is required to stop the automobile before it hits the barrier? 0.41 0.50 0.58 0.82 A system of two cables supports al50N weight as shown. What is the tension in the right hand cable? 87 N 150 N 173 N 260 N cannot be determined What is the tension in the horizontal cable? 87 N 150 N 173 N 300 N 260 N A spring scale is loosely fastened to the ceiling of a railway car. When a 1.0 kg mass is hung from the scale, it reads 12.0 N and is oriented as shown in the figure below What is the approximate acceleration of the car as measured by an observer on the ground? outside the car? 7.0 m/s^2 to the right 7.0 m/s^2 to the left 12.0 m/s^2 to the right 12 m/s^2 to the left It is impossible to calculate since the angle theta has not been given.

Explanation / Answer

20)

Total mass = 4kg + 2kg = 6 kg

acceleration = force / mass = 12/6 = 2 m/s^2

so force on 2kg block = mass*acceleration = 2kg * 2 m/s^2 = 4 N

The force exerted by the 4kg block on the 2kg block is equal to the force exerted by 2kg block on the 4kg block

but opposite in sign

ans - (b) -4N

21)

Let coefficient of kinetic friction be 'k'

mass be 'm'

friction force applied on the automobile will be k*m*g

initial velocity(U) = 20m/s

final velocity(V) = 0 [since cart should be stopped]

distance(s) = 50m

substituting the values in the formula " V^2 - U^2 = 2*a*s

a = -4 m/s^2

decceleration from friction force = force/mass = -k*m*g/m = -k*g

-k*g = -4

k = 4/9.8 = 0.41

ans - (a)

22)

let tension in the right hand cable be T1

T1 sin30 = 150

T1 = 150/sin30 = 300N

ans - e

23)

let tension in the horizontal cable be T2

T2 = T1 cos30

T2 = 300 * cos30

T2 = 260N

ans - d

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