1. The problem statement, all variables and given/known data A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. Find the work done by the force on the object as it moves as follows: (a) from x = 0 to x = 4.00 m (b) from x = 5.00 m to x = 10.0 m (c) from x = 11.0 m to x = 15.0 m. (d) If the object has a speed of 0.500 m/s at x = 0, find: speed at x = 5.00 m speed at x = 15.0 m 3. The attempt at a solution a. W=2.5N (4m) / 2 = 5J b. W=3N (5m) = 15J c. W=2.5N (4m) / 2 = 5J d. W=1/2 mv^{2} - 1/2 mvo^{2} x=5m 15J = 3kg v - (3kg)(0.500m/s) v=2.29 m/s x=15m 0=3kg v - (3kg)(0.500m/s) v=0.5 m/s
I would not use 2.5 N for the force. I suggest that you find an equation for the straight line in the form F = mx and calculate the force at 5 m. Looks good. Same comment as part (a) What happened to the 1/2 and the square of the speed in the expressions above?