A 3-kg object in pulled upward by a vertical cord (Use g^-10 m/s^2) What is the

Question

A 3-kg object in pulled upward by a vertical cord (Use g^-10 m/s^2) What is the tension in the cord when the velocity in constant at 4 m/s What tension-force is needed to accelerate the object from rest to 35 m/s in 5 N The cord breaks at r = 5s. What is the velocity 1.5 sec later (that is, at t = 6.5 s) A uniform horizontal rod of length 18 cm and weight 2 N is free to rotate about a hinge H at one end, while 4 N weight (text not clear) at the other end. The rod is supported by a vertical cord at the point 12 cm from the hinge. Taking the hinge as pivot, calculation the total gravitational torque. Find also the tension- force in the cord and the force exerted on the rod by the hinge. How far from the hinge must the cord be attached to make the hinge-force vanish The cord suddenly breaks, If I_TTOTAL = 0.015 kg m^2 (rod plus weight), what is the angular velocity 0.02 sec later The rod is now allowed to hang vertically from the same pivot. What is the period of small oscillations

Explanation / Answer

5.

a)

Forces acting on the object

T = tension in cord upward

mg = weight of object downward

at constant velocity , a =0

so force equation is given as

T - mg = ma

T - (3)(10) = 3 (0)

T = 30 N

b)

Vi = initial velocity = 0

Vf = final velocity = 3.5

a = acceleration

t = time taken = 5 sec

acceleration is given as

a = (Vf - Vi) / t = 3.5 - 0 /5 = 0.7 m/s2

so force equation is given as

T - mg = ma

T - (3)(10) = 3 (0.7)

T = 32.1 N

c)

Vi = 3.5 m/s

Vf = final velocity

a = acceleration = - 10

t = time taken = 1.5 s

using the equation

Vf = Vi + a t

Vf = 3.5 + (-10) (1.5)

Vf = - 11.5 m/s

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