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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 0 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? 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 (90✔ -(95.1 N) 5.11✓ 5.11 m) cos 109.5 110 90 19.5)-109.5 110 to the motion. The work done by gravity on the crate is given by Part 4 of 7-Analyze (b) We set the x and y axes parallel and perpendicular to the incline. From Newton's second law, we know that and we have n (95.1 N) cos 19.5✔ 19.5 son 89.7✔ Substituting for the coefficient of friction and the normal force, 0.4 89.6 Therefore, for the change in internal energy we have Part 5 of 7-Analyze (c) For the work done by the force on the crate, we have W-F 5.11 m) cos✔ 89.6 N. 35.9 N. 5.11 m)-103.35 183 1. 567 3

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 0 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? 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 (90✔ -(95.1 N) 5.11✓ 5.11 m) cos 109.5 110 90 19.5)-109.5 110 to the motion. The work done by gravity on the crate is given by Part 4 of 7-Analyze (b) We set the x and y axes parallel and perpendicular to the incline. From Newton's second law, we know that and we have n (95.1 N) cos 19.5✔ 19.5 son 89.7✔ Substituting for the coefficient of friction and the normal force, 0.4 89.6 Therefore, for the change in internal energy we have Part 5 of 7-Analyze (c) For the work done by the force on the crate, we have W-F 5.11 m) cos✔ 89.6 N. 35.9 N. 5.11 m)-103.35 183 1. 567 3

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Author:William Moebs, Samuel J. Ling, Jeff Sanny
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Chapter8: Potential Energy And Conservation Of Energy
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Hello I help with (c) only in the second image, did part (a), (b), and (c) already.
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 0 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? 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 (90✔ -(95.1 N) 5.11 5.11 m) cos 109.5 110 90 19.5) 109.5 110 to the motion. The work done by gravity on the crate is given by Part 4 of 7-Analyze (b) We set the x and y axes parallel and perpendicular to the incline. From Newton's second law, we know that and we have (95.1 N) cos 19.5✔ 19.5 son 9.7✔ 89.6 N. Substituting for the coefficient of friction and the normal force, 0.4 89.6 Therefore, for the change in internal energy we have 35.9✔ 35.9 (5.11✔ Part 5 of 7-Analyze (c) For the work done by the force on the crate, we have W-F 5.11 m) cos✔ 35.9 N. 5.11 m)-183.35 183 J. 567 3
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 0 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? 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 (90✔ -(95.1 N) 5.11 5.11 m) cos 109.5 110 90 19.5) 109.5 110 to the motion. The work done by gravity on the crate is given by Part 4 of 7-Analyze (b) We set the x and y axes parallel and perpendicular to the incline. From Newton's second law, we know that and we have (95.1 N) cos 19.5✔ 19.5 son 9.7✔ 89.6 N. Substituting for the coefficient of friction and the normal force, 0.4 89.6 Therefore, for the change in internal energy we have 35.9✔ 35.9 (5.11✔ Part 5 of 7-Analyze (c) For the work done by the force on the crate, we have W-F 5.11 m) cos✔ 35.9 N. 5.11 m)-183.35 183 J. 567 3
Part 6 of 7 - Analyze (d) We use the energy version of the nonisolated system model to find the change in kinetic energy. ΔΚ = ΣWother =WF + Wg ΔΕ AE int AE int J J+ J J Note that the normal force does zero work because it is at o to the motion.
Transcribed Image Text:Part 6 of 7 - Analyze (d) We use the energy version of the nonisolated system model to find the change in kinetic energy. ΔΚ = ΣWother =WF + Wg ΔΕ AE int AE int J J+ J J Note that the normal force does zero work because it is at o to the motion.
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