An object with a mass of #4 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 5+tanx #. How much work would it take to move the object over #x in [(-5pi)/12, (5pi)/12], where x is in meters?

Question

An object with a mass of #4 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 5+tanx   #.  How much work would it take to move the object over #x in [(-5pi)/12, (5pi)/12], where x is in meters?

Julia Aguilar · Accepted Answer

The work is #=513.1J#

The mass is #m=4kg#.

#F_r=mu_k*mg#

#=4(5+tanx)g#

The completed work is

#W=4gint_(-5/12pi)^(5/12pi)(5+tanx)dx#

[5x+ln|cos(x)|]_(-5/12pi)^(5/12pi)# = 4g*

#=4g((5*5/12pi+ln|cos(5/12pi)|)-(5*-5/12pi+ln|cos(-5/12pi)|))#

#=4g(50/12pi+ln|cos(5/12pi)|-ln|cos(5/12pi)|)#

#=4g(25/6pi+0)#

#=513.1J#

Nathan Ashcroft · Answer

undefined $$W = \int_{-5\pi/12}^{5\pi/12} \left(\mu_k(x) \cdot m \cdot g + \frac{m \cdot v^2(x)}{2}\right) \, dx$$

$$W = \int_{-5\pi/12}^{5\pi/12} \left((5 + \tan(x)) \cdot 4 \cdot 9.8 + \frac{4 \cdot v^2(x)}{2}\right) \, dx$$

Axel Acevedo · Answer

undefined To find the work done to move the object over the given interval, we integrate the force of kinetic friction with respect to displacement (x) over the specified range. The work done by friction is given by the formula:

$$ W = \int_{x_1}^{x_2} f_k(x) \, dx $$

Given that the coefficient of kinetic friction $ \mu_k(x) = 5 + \tan(x) $, we can find the force of kinetic friction $ f_k(x) $ using the formula:

$$ f_k(x) = \mu_k(x) \cdot N $$

where N is the normal force. Since the object is pushed along a linear path, the normal force N is equal to the gravitational force acting on the object, which is $ mg $, where $ m = 4 \, \text{kg} $ is the mass of the object and $ g = 9.8 \, \text{m/s}^2 $ is the acceleration due to gravity.

So, $ N = mg = 4 \times 9.8 = 39.2 \, \text{N} $.

Therefore, $ f_k(x) = (5 + \tan(x)) \times 39.2 $.

Now, we integrate $ f_k(x) $ with respect to $ x $ over the interval $ \left[ \frac{-5\pi}{12}, \frac{5\pi}{12} \right] $:

$$ W = \int_{-\frac{5\pi}{12}}^{\frac{5\pi}{12}} (5 + \tan(x)) \cdot 39.2 \, dx $$

After finding this integral, you'll have the work done to move the object over the specified interval.