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Change in Kinetic energy

  • Thread starter Joshua Lee
  • Start date
1. Homework Statement
A.What is the change in kinetic energy of the crate? Answer in units of J.
B.What is the speed of the crate after it is pulled 5.79 m? Answer in units of m/s
Magnitude of work done by gravity= 200.786
Work done by applied force= 822.18
mass=8.91kg
force pulling on box= 142N
angle from horizontal to force= 23.4 degrees
crate is moving at a speed of 1.48 m/s
coefficient of friction= 0.292

2. Homework Equations
Change in Kinetic Energy= Work done by gravity + work done by the applied force

KE= (1/2)mv2

3. The Attempt at a Solution
A. 822.18-200.786(since gravity is negative)= 621.394J
B. Found Fn=mgcos(theta)
Multiplied that by the coefficient of friction. Got the work done by friction. Subtracted that from work done by the applied force. Made that KE and used the second equation to solve for v which I got as 5.1.
 

haruspex

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It is unclear what you are originally given and what you have deduced. Please start with a complete statement of the problem as presented to you, including either a diagram or a good verbal description of the set-up.
 

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