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

Question

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

Layla Ahmed · Accepted Answer

The work done is #=8J# The mass is #m=2kg# The exerted force is #F(x)=x+3# The completed work is #W=F*d# #d# is the distance Thus, #W=intF(x)dx# #=int_0^2(x+3)dx# #=[1/2x^2+3x]_0^2# #=(2+6)-(0)# #=8J#

Charles Aeschliman · Answer

undefined To find the work done in moving the object from $ x = 0 $ to $ x = 2 $, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

The work done ($ W $) is given by the integral of the force function $ F(x) $ with respect to displacement $ x $:

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

Substituting the given force function $ F(x) = x + 3 $ and the limits of integration $ x_1 = 0 $ and $ x_2 = 2 $, we get:

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

$$ W = \left[ \frac{x^2}{2} + 3x \right]_{0}^{2} $$

$$ W = \left( \frac{2^2}{2} + 3 \times 2 \right) - \left( \frac{0^2}{2} + 3 \times 0 \right) $$

$$ W = \left( 2 + 6 \right) - \left( 0 + 0 \right) $$

$$ W = 8 \text{ joules} $$

Therefore, it would take 8 joules of work to move the object over the interval [0, 2].