Home > Physics > Frictional Forces

An object with a mass of #5 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= e^x+x #. How much work would it take to move the object over #x in [3, 4], where x is in meters?

Answer 1

Charlotte Aaron

The work done is #=1862.5J#

#"Reminder : "#

#inte^xdx=e^x+C#

The completed work is

#W=F*d#

The force of friction is

#F_r=mu_k*N#

The coefficient of kinetic friction is #mu_k=(e^x+x)#

The normal force is #N=mg#

The mass of the object is #m=5kg#

#F_r=mu_k*mg#

#=5*(e^x+x)g#

The completed work is

#W=5gint_(3)^(4)(e^x+x)dx#

#=5g*[e^x+x^2/2]_(3)^(4)#

#=5g((e^4+8)-(e^3+9/2))#

#=5g(38.01)#

#=1862.5J#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Olivia Askew

To calculate the work done, use the integral of the friction force over the given interval:

[ W = \int_{3}^{4} u_k(x) \cdot m \cdot g ,dx ]

where ( m ) is the mass (5 kg), ( g ) is the acceleration due to gravity (approximately 9.8 m/s²).

[ W = \int_{3}^{4} (e^x + x) \cdot 5 \cdot 9.8 ,dx ]

Approximately, the work done is 1565.78 Joules.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.