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How much work would it take to horizontally accelerate an object with a mass of #6 kg# to #7 m/s# on a surface with a kinetic friction coefficient of #3 #?

Answer 1

Adrian Ayala

The work is #=147J#

The coefficient of kinetic friction is

#mu_k=F_r/(N)#

The normal force is

#(N)=mg=6g N#

The frictional force is

#F_r=mu_k*(N)=3*6g=18gN#

According to Newton's Second Law, the acceleration is

#F=ma#

#a=F/m=18g/6=3g#

To find the distance, we apply the equation of motion

#v^2=u^2+2as#

#u=0#

#s=v^2/(2a)=7^2/(2*3g)m#

The work is

#W=Fs=18g*49/(6g)=147J#

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Answer 2

Elena Albarran

To horizontally accelerate an object with a mass of 6 kg to a velocity of 7 m/s on a surface with a kinetic friction coefficient of 0.3, it would require approximately 103.86 Joules of work.

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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.