Question 1 A horizontal force of magnitude 46.5 N pushes a block of mass 3.91 kg

Question

Question 1 A horizontal force of magnitude 46.5 N pushes a block of mass 3.91 kg across a floor where the coefficient of kinetic friction is 0.612. (a) How much work is done by that applied force on the block-floor system when the block slides through a displacement of 2.91 m across the floor? (b) During that displacement, the thermal energy of the block increases by 38.1 what is the increase in the rmal energy of the floor? (c) what is the increase in the kinetic energy of the block? (a) Number (b) Number Units Units Units (c) Number

Explanation / Answer

a)

We know that work done by a force is

W=F.d

where F is the applied force and d is the displacment.

So ,

when the block moves 2.91m , work done will be,

W=46.5*2.91

=135.315J

b)

We know that total energy of the system(block+floor) will be conserved as energy can niether be created nor be

destroyed.

So,

change in the kinetic energy of the block will,

net horizontal force on the block is,

(46.5-.612*3.91*9.8)

23.0493N

so net horizontal acceleration will be..F/m

=5.8949

Now,

v2-u2=2as

so change in kinetic energy of the block is

=mas

3.91*5.8949*2.91

=67.07370J(increase in kinetic energy of block)......part(c)

so

increase in thermal energy of floor+increase in thermal energy of block=increase in kinetic energy of block.

SO,

increase in thermal energy of floor=increase in kinetic energy of block-increase in thermal energy of block

=67.0737-38.1

=28.973J

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