Chapter 3: Q3.4-19E (page 116)

An object of mass 60 kg starts from rest at the top of a 45º inclined plane. Assume that the coefficient of kinetic friction is 0.05 (see Problem 18). If the force due to air resistance is proportional to the velocity of the object, say, -3, find the equation of motion of the object. How long will it take the object to reach the bottom of the inclined plane if the incline is 10 m long?

Short Answer

Expert verified

The equation of motion of an object is $x (t) = 131 . 8 (t + 20 e^{- 1 / 20}) - 2636$
The object take to reach the bottom of inclined plan is 1.768 seconds,if the incline is 10 m

Step by step solution

Find the equation of motion of an object

There are two forces are

$F_{1} = mg \sin 45^{o} F_{2} = - μ mg \cos 45^{o} F_{3} = - 3 v$

Now

$m \frac{dv}{dt} = mg \sin 45^{o} - μ mg \cos 45^{o} - 3 v \frac{dv}{dt} = 6.57 - \frac{v}{20} \int \frac{20 dv}{131.8 - v} = \int dt \int \frac{20 dv}{131.8 - v} = t + C$

Find ∫20 dv131.8-v.

Put $| 131.8 - v = t dv = - dt |$ then

$\int \frac{20 dv}{131.8 - v} = - 20 \int \frac{dt}{t} = - 20 \ln |t| = - 20 \ln |131.8 - v| = - 20 \ln |131.8 - v| = t + C = \ln |131.8 - v| = - \frac{t}{20} + C v (t) = 131 . 8 - {Ce}^{- 1 / 20}$

When v(t_o) = 0 then C = 131.8

$v (t) = 131 . 8 (1 - e^{\frac{- 1}{20}})$

Find the value of x(t)

$x (t) = \int 131 . 8 (1 - e^{\frac{- 1}{20}}) dt x (t) = 131 . 8 (t + 20 e^{\frac{- 1}{20}}) + C$

When x(0) = 0 then C=-2636 so

$x (t) = 131 . 8 (t + 20 e^{\frac{- 1}{20}}) - 2636$

Hence, the equation of motion of an object is $x (t) = 131 . 8 (t + 20 e^{- 1 / 20}) - 2636$

Find the value of time

When x(t) = 10 then $10 = 131 . 8 (t + 20 e^{\frac{- 1}{20}}) - 2636$

By solving it, we obtain t = 1.768 sec.

Hence, the object takes to reach the bottom of inclined plan is 1.768 seconds, if the incline is 10 m.

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Short Answer

Step by step solution

Find the equation of motion of an object

Find ∫20 dv131.8-v.

Find the value of x(t)

Find the value of time

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