A crate of mass m = 9.7 kg is pulled up a rough incline with an initial speed of v; = 1.43 m/s. The pulling force is F = 111 N parallel to the incline, which makes an angle of = 19.5° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled d = 5.11 m. (a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate-incline system owing to friction. (c) How much work is done by the 111-N force on the crate? (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5.11 m?

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need help with last part only solve for (a) the force of gravitation is full part.
A crate of mass m = 9.7 kg is pulled up a rough incline with an initial speed of v; = 1.43 m/s. The pulling force is F = 111 N parallel to the incline, which makes an angle of = 19.5° with the horizontal. The
coefficient of kinetic friction is 0.400, and the crate is pulled d = 5.11 m.
(a) How much work is done by the gravitational force on the crate?
(b) Determine the increase in internal energy of the crate-incline system owing to friction.
(c) How much work is done by the 111-N force on the crate?
(d) What is the change in kinetic energy of the crate?
(e) What is the speed of the crate after being pulled 5.11 m?
mg
Part 1 of 7 - Conceptualize
The gravitational force does negative work of some tens of joules on the crate. We expect some hundreds of joules of work done by the force on the crate. This force should be larger than the increase in the
internal energy of the system and larger than the change in kinetic energy of the crate. We estimate the final speed to be a few meters per second.
Part 2 of 7 - Categorize
We could use Newton's second law to find the crate's acceleration, but using ideas of work and energy is a more direct way to calculate the final speed. Finding the increase in internal energy is a step towards
finding the temperature increase of the rubbing surfaces.
Part 3 of 7 - Analyze
(a) The force of gravitation is (9.7 kg) (9.80 m/s²) = 95.1 N straight down, at an angle of
W₂ =F. AF
-(95.1 N
m cos
J.
° +
19.5°).
=
to the motion. The work done by gravity on the crate is given by
Transcribed Image Text:A crate of mass m = 9.7 kg is pulled up a rough incline with an initial speed of v; = 1.43 m/s. The pulling force is F = 111 N parallel to the incline, which makes an angle of = 19.5° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled d = 5.11 m. (a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate-incline system owing to friction. (c) How much work is done by the 111-N force on the crate? (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5.11 m? mg Part 1 of 7 - Conceptualize The gravitational force does negative work of some tens of joules on the crate. We expect some hundreds of joules of work done by the force on the crate. This force should be larger than the increase in the internal energy of the system and larger than the change in kinetic energy of the crate. We estimate the final speed to be a few meters per second. Part 2 of 7 - Categorize We could use Newton's second law to find the crate's acceleration, but using ideas of work and energy is a more direct way to calculate the final speed. Finding the increase in internal energy is a step towards finding the temperature increase of the rubbing surfaces. Part 3 of 7 - Analyze (a) The force of gravitation is (9.7 kg) (9.80 m/s²) = 95.1 N straight down, at an angle of W₂ =F. AF -(95.1 N m cos J. ° + 19.5°). = to the motion. The work done by gravity on the crate is given by
(a) The force of gravitation is (9.7 kg)(9.80 m/s²) = 95.1 N straight down, at an angle of
W₁ =F-AF
(95.1 N
m cos
J.
19.5°)
to the motion. The work done by gravity on the crate is given by
Transcribed Image Text:(a) The force of gravitation is (9.7 kg)(9.80 m/s²) = 95.1 N straight down, at an angle of W₁ =F-AF (95.1 N m cos J. 19.5°) to the motion. The work done by gravity on the crate is given by
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