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Physics For Scientists & Engineers
Found in: Page 238

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Short Answer

The coefficient of friction between the block of mass m1=3.00 kgand the surface in Figure P8.22 is.μk=0.400 The system starts from rest. What is the speed of the ball of mass m2=5.00 kgwhen it has fallen a distanceh=1.5 m?

The speed of the ball when it has fallen a distance h=1.5 m isv=3.74  m/s .

See the step by step solution

Step by Step Solution

Step 1: Law of conservation of energy

In the presence of dissipative forces, the law of conservation of energy states that the total change in mechanical energy during a process, should be equal to the increase in internal energy of the system.

ΔEmech=ΔK+ΔUg=-fkd=-ΔEint  ...............(8.16)

If a system is not isolated as well, the total change in energy is given as-

ΔE=ΔK+ΔU+ΔEint=Wother forces   ............(8.17)


ΔK= Change in kinetic energy

ΔU= Change in potential energy

ΔEint= Change in internal energy

Step 2: Given Data

mass of the block: m1=3.00 kg

mass of the ball: m2=5.00 kg

coefficient of friction: μk=0.400

distance moved: h=1.5 m

Step 3:

Use Equation 8.16:


For the system of block, ball and surface; let the subscript ‘1’ and ‘2’ denote the values of block and ball respectively. So, equation becomes-

data-custom-editor="chemistry" ΔK1+ΔU1+ΔK2+ΔU2=fkd

Choose the initial point before release and the final point after each block have moved1.5 m . Chooseu=0 with the3.00 kg block on the tabletop and the block in its final position.

So Ki1=Ki2=U1f=U2f=Ui1=0

We now have


where the friction force is


The friction force causes a negative change in mechanical energy because the force opposes the motion. Since all of the variables are known except for,vf we can substitute and solve for the final speed.


v2=2gh(m2μkm1)m1+m2v=2(9.8 m/s2)(1.5 m)[5.00 kg-0.400(3.00 kg)]8.00 kgv=3.74 m/s

After falling 1.5 m, the ball is moving with a velocity of .3.74 m/s

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