2.40X TO IN TorcC 13 pu m The force acts on an angle that is 20.0° above the horizontal. kinetic friction is 0.200. After the refrigerator moves for 8.0 work done by the pulling force. What is the work done by the he refrigerator is initially at rest, use the work-kinetic energy inal speed of the refrigerator.
2.40X TO IN TorcC 13 pu m The force acts on an angle that is 20.0° above the horizontal. kinetic friction is 0.200. After the refrigerator moves for 8.0 work done by the pulling force. What is the work done by the he refrigerator is initially at rest, use the work-kinetic energy inal speed of the refrigerator.
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![A 2.40×10²N force is pulling an 85.0kg refrigerator across a horizontal surface.
The force acts on an angle that is 20.0° above the horizontal. The coefficient of
kinetic friction is 0.200. After the refrigerator moves for 8.00 meters, find the
work done by the pulling force. What is the work done by the kinetic friction? If
the refrigerator is initially at rest, use the work-kinetic energy model to find the
final speed of the refrigerator.
Answer: the work done by the pulling force will be 1.80×10°J. The work done
by the kinetic friction will be 1.20×10'J. The net work done will be
6.01×10²J. As a result, the final speed will be v=3.76ms.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F369dfef0-c767-43fc-ad35-2d7631c460d1%2Fa28fc936-0cab-4878-b20d-55c0c5638cb6%2F46cac9n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A 2.40×10²N force is pulling an 85.0kg refrigerator across a horizontal surface.
The force acts on an angle that is 20.0° above the horizontal. The coefficient of
kinetic friction is 0.200. After the refrigerator moves for 8.00 meters, find the
work done by the pulling force. What is the work done by the kinetic friction? If
the refrigerator is initially at rest, use the work-kinetic energy model to find the
final speed of the refrigerator.
Answer: the work done by the pulling force will be 1.80×10°J. The work done
by the kinetic friction will be 1.20×10'J. The net work done will be
6.01×10²J. As a result, the final speed will be v=3.76ms.
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