The force applied against a moving object travelling on a linear path is given by #F(x)=3x^2+e^x #. How much work would it take to move the object over #x in [0, 3 ] #?

Question

The force applied against a moving object travelling on a linear path is given by #F(x)=3x^2+e^x #. How much work would it take to move the object over #x in [0, 3 ]   #?

Elena Albarran · Accepted Answer

The work is #=46.1J#

We require

#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#

#inte^xdx=e^x+C#

The task is

#DeltaW=FDeltax#

#DeltaW=(3x^2+e^x)Deltax#

#W=int_0^3(3x^2+e^x)dx#

#=[x^3+e^x]_0^3#

#=(3^3+e^3)-(0+e^0)#

#=27+e^3-1#

#=46.1J#

Charlotte Aaron · Answer

I got #46J#.

Here you have a variable force so for an infinitesimal (=very small) Work you have: #dW=F(x)dx# integrating: #W=int_(x_1)^(x_2)F(x)dx# in your situation: #W=int_(0)^(3)(3x^2+e^x)dx=cancel(3)x^3/cancel(3)+e^x=x^3+e^x|_0^3=# Utilize the Calculus Fundamental Theorem: #=(3^3+e^3)-(0^3+e^0)=46J#

Adrian Ayala · Answer

undefined To calculate the work done, you need to integrate the force function F(x) with respect to x over the interval [0, 3]:

∫[0 to 3] (3x^2 + e^x) dx

The definite integral of 3x^2 + e^x with respect to x from 0 to 3 is approximately 206.5. Therefore, it would take approximately 206.5 units of work to move the object over the interval [0, 3].

Layla Ahmed · Answer

undefined To find the work done to move the object over the interval $[0, 3]$, we need to integrate the force function $F(x) = 3x^2 + e^x$ with respect to $x$ over the given interval. The work done is given by the integral:

$$ W = \int_{0}^{3} F(x) \, dx $$

$$ W = \int_{0}^{3} (3x^2 + e^x) \, dx $$

Integrate each term separately:

$$ W = \left[ x^3 + e^x \right]_{0}^{3} $$

$$ W = (3^3 + e^3) - (0^3 + e^0) $$

$$ W = (27 + e^3) - (1) $$

$$ W = 27 + e^3 - 1 $$

$$ W = 26 + e^3 $$