An object with a mass of #1 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= e^x-x+3 #. How much work would it take to move the object over #x in [1, 2], where x is in meters?

Question

An object with a mass of #1 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= e^x-x+3   #.  How much work would it take to move the object over #x in [1, 2], where x is in meters?

Charles Aeschliman · Accepted Answer

The work done is #=65.7J#

We need

#inte^xxdx=e^x#

The frictional force is

#F=F_r=mu_k*N#

The normal force is #N=mg#

So,

#F_r=mu_k*mg#

#m=1kg#

But,

#mu_k(x)=e^x-x+3#

The work done is

#W=F_r*d=int_0^(pi/12)mu_kmgdx#

#=gint_1^2(e^x-x+3)dx#

#=g[e^x-x^2/2+3x]_1^2#

#=g((e^2-2+6)-(e-1/2+3))#

#=g*(e^2-e+3/2)#

#=6.17g#

#=65.7J#

Parker Abbott · Answer

undefined To calculate the work done to move the object over the interval [1, 2], we need to integrate the force of kinetic friction over that interval. The force of kinetic friction $ F_k $ is given by the equation:

$$ F_k = \mu_k \cdot N $$

where $ \mu_k $ is the kinetic friction coefficient and $ N $ is the normal force. Since the object is being pushed along a linear path, the normal force is equal to the weight of the object, which is $ mg $, where $ m $ is the mass of the object and $ g $ is the acceleration due to gravity.

The work done $ W $ is then given by the equation:

$$ W = \int_{1}^{2} F_k \, dx $$

Substituting $ F_k = \mu_k(x) \cdot mg $ into the equation, we get:

$$ W = \int_{1}^{2} \left( e^x - x + 3 \right) \cdot mg \, dx $$

Given that the mass of the object is $ m = 1 \, \text{kg} $, and $ g = 9.8 \, \text{m/s}^2 $, we can compute the integral:

$$ W = \int_{1}^{2} \left( e^x - x + 3 \right) \cdot 1 \cdot 9.8 \, dx $$

$$ W = \left[ 9.8 \cdot \left( e^x - \frac{x^2}{2} + 3x \right) \right]_{1}^{2} $$

$$ W = \left[ 9.8 \cdot \left( e^2 - \frac{2^2}{2} + 3 \cdot 2 - e - \frac{1^2}{2} + 3 \cdot 1 \right) \right] $$

$$ W = 9.8 \cdot \left( e^2 - 2 + 6 - e - \frac{1}{2} + 3 \right) $$

$$ W = 9.8 \cdot \left( e^2 - e + \frac{11}{2} \right) $$

$$ W \approx 9.8 \cdot \left( 7.389 - 2.718 + 5.5 \right) $$

$$ W \approx 9.8 \cdot 10.171 $$

$$ W \approx 99.84 \, \text{Joules} $$

Therefore, it would take approximately $ 99.84 \, \text{Joules} $ of work to move the object over the interval [1, 2] meters.