1. (Mon) You are pushing a 500 kg car that ran out of gas and is at rest on a flat road. If you apply a constant force for 30 seconds and move the car 50 m, what force did you apply to the car (neglect friction)? (8 min / 8 pts)

2. Δx = v i t + ½at ² (Mon) You are pushing a 500 kg car that ran out of gas and is at rest on a flat road. If you apply a constant force for 30 seconds and move the car 50 m, what force did you apply to the car (neglect friction)? (8 min / 8 pts) F = 55.56 N 50 m = 0 + (a)(450 s ² ) 50 m/450 s ² = a a = 0.11 m/s² 50 m = (0m/s)(30s)+(0.5)(a )(30s)² xy ΔxΔx vivi vfvf a t m N/A 50 30 ??? 0 500 F = ma 0.11 F = maF = (500kg)(0.11 m/s ²)

3. (Tue) You push a 500 kg car that ran out of gas and is at rest on a flat road. If you apply a constant force for 30 seconds and move the car 50 m, how much work did you do on the car (neglect friction)? (8 min / 8 pts)

4. Δx = v i t + ½at ² (Tue) You push a 500 kg car that ran out of gas and is at rest on a flat road. If you apply a constant force for 30 seconds and move the car 50 m, how much work did you do on the car (neglect friction)? (8 min / 8 pts) W = 2778 J 50 m = 0 + (a)(450 s ² ) 50 m/450 s ² = a a = 0.11 m/s² 50 m = (0m/s)(30s)+(0.5)(a )(30s)² xy ΔxΔx vivi vfvf a t m N/A 50 30 ??? 0 500 F = ma 0.11 F = maF = (500kg)(0.11 m/s ²) W = FdF = 55.56 N W = (55.56 N)(50 m)

5. (Wed) You push a 500 kg car that is at rest on a flat road. If the μ k between the car and the road is 0.25, you apply a constant force for 30 s and move the car 50 m, how much force did you apply to the car? (10 min/8 pts)

6. 50m = 0 + (0.5)(a)(30s) ² F g = ma g F g = (500kg)(9.81m/s ²) (Wed) You push a 500 kg car that is at rest on a flat road. If the μ k between the car and the road is 0.25, you apply a constant force for 30 s and move the car 50 m, how much force did you apply to the car? (10 min/8 pts) F p = 1282 N F n = F g F n = 4905 N F g = 4905 N xy ΔxΔx vivi vfvf a t m N/A 50 30 ??? 0 500 0.11 F f = F n μ k F f = (4905)(0.25) Find F net in ‘x’ direction F f = 1226.25 N Δx = v i t + ½at ² xy FgFg FnFn FfFf FpFp F net N/A 0 4905 1226 0.11m/s ² = aF net = ma F net = (500kg)(0.11 m/s ²) F net = 55.6 N 55.6 1282 F net = F p - F f

7. (Thu) A 3.00 kg block starts from rest at the top of a 30.0 º incline and accelerates uniformly down the incline moving 2.00 m in 1.50 s. If the net force down the ramp is 5.34 N, what is the force due to friction acting on the block? (5 min / 5 pts)

8. x Diry Dir FgFg FnFn FfFf F net 14.71 N 0 N 5.34 N F g,x = ma g sin(30) F g,x = (3.00 kg)(9.81 m/s²)(0.5) F g,x = 14.71 N F net = F g + F n + F f 5.34 N = 14.71 N + 0 + F f -9.37 N = F f ???? In the ‘x’ direction…..

9. (Fri) A baseball is pitched at home plate with a speed of 35 m/s. If the baseball has a mass of 150 g, what is the ball’s Kinetic Energy? (5 min / 5 pts)

10. KE = 91.9 J KE = ½mv ² (Fri) A baseball is pitched at home plate with a speed of 35 m/s. If the baseball has a mass of 150 g, what is the ball’s Kinetic Energy? (5 min / 5 pts) KE = (0.5)(0.150 kg)(35 m/s) ² KE = 91.875 kgm ²/s² And 1 just ‘cause….

11. End of Week Procedures: 1.Add up all the points you got this week 2.Put the total number of points at the top of your page (out of 34 points) 3.List any dates you were absent (and why) 4.Make sure your name and period is on the top of the paper 5.Turn in your papers