Holt

1
Holt

• Problem 5 p. 176

2
State the problem

• A 10.0 kg crate is pulled up a rough incline with
  an initial speed of 1.5 m/s. The pulling force
  is 100.0 N parallel to the incline, which makes
  an angle of 15.0o with the horizontal. Assuming
  the coefficient of kinetic friction is 0.40 and
  the crate is pulled a distance of 7.5 m, find the
  following

3
State the problem

• a. The work done by the Earths gravity on the
  crate.
• b. The work done by the force of friction on the
  crate.
• c. The work done by the puller on the crate.
• d. The change in the kinetic energy of the
  crate.
• e. The speed of the crate after it is pulled 7.5
  m.

4
Make a drawing
100 N
vi1.5 m/s
5
Make a drawing
What is the final velocity after moving the 7.5 m?
vi1.5 m/s
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Complete the f.b.d
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Start solving

• What is the work done by the earths gravity on
  the crate?
• Work Force x Distance
• Work - mg sin ? x distance
• Work - (10.0 kg)(9.81 m/s2)sin 15.00 x 7.5 m
• Work - 190 J

Notice negative sign
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b. What is the work done by the force of
friction on the crate? Work Force x
Distance Work - Kinetic friction x
distance Work - (µk x FN) x distance so what
is FN? mg cos ? Work - (0.40 x (10.0
kg)(9.81 m/s2) cos 15.00) x 7.5 m Work - 284
J
Notice negative sign
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c. What is the work done by the puller on the
crate? Work Force x Distance Work 100 N x 7.5
m Work 750 J
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d. What is the change in kinetic energy of the
crate? From the work-kinetic energy
theorem WorkNET ?KE KEF KEi From parts a,
b, c WorkNET (- 190 J) (- 284 J) (750
J) WorkNET 276 J Therefore ?KE 276 J
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• What is the speed of the crate after it is pulled
  7.5 m?
• Again, from the Work Kinetic Energy Theorem
• WorkNET ?KE KEF KEi
• From part d ?KE 276 J
• ?KE KEF KEi ½ mvf2 ½ mvi2 ?KE
• Solve for vf
• ½ mvf2 ?KE ½ mvi2
• vf 2 2(?KE ½ mvi2)/m
• vf 2 2(276J ½ x 10 kg x(1.5 m/s) 2)/10 kg
• vf 2 57.45
• vf 7.6 m/s