1. The problem statement, all variables and given/known data 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. Relevant 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.