On a flat surface a 25kg crate is pushed at a constant velocity with a force of 50N. Determine the coefficient of friction. What would the acceleration on the crate be if the applied force was doubled?

and how far would the crate move in the 5 seconds if it began with a velocity of 2m/s?

1 Answer
Michael
Nov 2, 2016

#sf((a))#

#sf(mu=0.2)#

#sf((b))#

#sf(a=2.0color(white)(x)"m/s"^(2))#

#sf((c))#

#sf(s=35color(white)(x)"m")#

Explanation:

#sf((a))#

Since the crate is moving at constant velocity the net force on the crate is zero.

The frictional force due to the surface is given by:

#sf(F=muR)#

Since the crate is moving we are referring to the coefficient of dynamic friction.

#sf(R)# is the reaction and is equal to the weight of the block mg.

#:.##sf(R=mg=25xx9.8=245color(white)(x)N)#

#sf(muRstackrel(color(white)(xxxxxxxxxxxxxx))(color(blue)(larr)#X#stackrel(color(white)(xxxxxxxxxxxxxx))(color(blue)(rarr)##sf(50"N")#

The frictional force is balanced by the effort so:

#sf(50=muxx245)#

#:.##sf(mu=50/245=0.2)#

#sf((b))#

The effort is now doubled so the forces on the block are not balanced.

#sf(muRstackrel(color(white)(xxxxxxxxxxxxxx))(color(blue)(larr)#X#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(rarr)##sf(100"N")#

The net force is given by:

#sf(F=100-muR)#

#:.##sf(F=100-(245xx0.2)=51color(white)(x)N)#

Using Newton's Law:

#sf(F=ma)#

#:.##sf(a=F/m=51/25=2.0color(white)(x)"m/s"^(2))#

#sf((c))#

I will assume you mean how far the crate moves after the force has been doubled.

Using the equation of motion:

#sf(s=ut+1/2at^(2))#

Putting in the numbers:

#sf(s=(2xx5)+1/2xx2xx5^(2)=35color(white)(x)m)#

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