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Chapter 3: Q9E (page 116)

An object of mass 100kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the object down, a buoyancy force of 1/40 times the weight of the object is pushing the object up (weight = mg). If we assume that water resistance exerts a force on the object that is proportional to the velocity of the object, with proportionality constant 10N-sec/m , find the equation of motion of the object. After how many seconds will the velocity of the object be 70 m/sec ?

Short Answer

Expert verified

The result is x鈥�(t)=95.65t-956.5e-t10-956.5 and time taken by the object is 13.2 sec .

Step by step solution

01

Important hint.

Use Newton鈥檚 method to solve for t tn+1=tn-f(tn)f'(tn).

02

Find the equation of motion

The total force acting on the object is F-mg-bv-140mg.

Applying newton鈥檚 second law:

100dvdt=100鈥�(9.81)-10鈥�v-1鈥�鈥�(100)鈥�鈥�(9.81)40

v'=9.56-0.1vv'+0.1v=9.56v'+0.1v=9.56鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�onsolvingbyvariableseparating

When v鈥�(0)=0,鈥�鈥�then鈥�鈥�C=-95.65.

v=95.65-95.65e-t10x(t)=95.65t-956.5e-t10+c 鈥︹�� (1)

When x(0)=0,鈥�鈥�then鈥�鈥�C=-956.5.

x鈥�(t)=95.65鈥�t-956.5鈥�e-t10-956.5

03

Find the value of t

When the object travelling at the velocity 70m/sec then by equation (1).

70=95.65鈥�t-956.5鈥�e-t1070=95.65(1-e-t10)t=13.2鈥�sec

Therefore, the result is x鈥�(t)=95.65鈥�t-956.5鈥�e-t10-956.5 and t=13.2鈥�sec.

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