A constant force of 8.0 N is exerted for 4.0 s on 16-kg object initially at rest

Question

A constant force of 8.0 N is exerted for 4.0 s on 16-kg object initially at rest. The change in speed of this object will be: 0.5 m/s 2 m/s 4 m/s 8 m/s 32 m/s A 0.2-kg stone is attached to a string and swung in a circle of radius 0.6 m on a horizontal and frictionless surface. If the stone makes 150 revolutions per minute, the tension force of the sting is: 0.03N 0.2N 0.9N 1.96N 30N An object of mass 1 g is whirled in a horizontal circle of radius 0.5 m at constant speed of 2 m/s. The work done on the object during one revolution is: 1J 0J 4J 2J 16J A car travels at a constant speed of 30mph (13.4 m/s) on a level circular turn of radius 50 m. If friction provides the necessary force to keep the car on the circular path, what is the minimum coefficient of friction between the tires and the roadway in order for the car to make the circular turn without sliding? The mass of the car is 1000 kg. A ball of mass 8.0 kg is suspended by two wires from a horizontal arm, which is attached to a vertical shaft, as shown in figure. The shaft is in uniform rotation about its axis such that the linear speed of the ball equals 2.3 m/s. Find the tensions in wires 1 and 2.

Explanation / Answer

(12) Acceleration, a = 8 / 16 = 0.50 m/s^2

Final velocity, v = 0 + 0.50*4 = 2 m/s.

Option (B) is the correct answer.

(13) w = 150 rev/m = 150/60 rev/s = 2.5 rev/s = 2.5*6.28 = 15.7 rad/s.

Tensile force in the string, F = mv^2/r = mw^2r = 0.2*15.7^2*0.6 = 29.58 N = 30 N

Option (E) is the correct answer.

(14) Displacement is ZERO. So, work done = 0

Option (B) is the correct answer.

(15) Frictional force = u*mg

This force shall be balanced by = mv^2/R

Now equate both -

u*mg = mv^2/R

=> u = v^2 / (R*g) = 13.4^2 / (50*9.8) = 0.366

So, coefficient of friction = 0.366

(16) Centrifugal force = mv^2/R

Now equate this force with T1 and T2, the tension in the two strings. By dividing horizontal and vertical components, you will get the results.

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