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Δ x = 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).

jelani-underwood
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Δ x = v i t + ½at ²

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  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. (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 = ma Δx = vit + ½at² 50 m = (0m/s)(30s)+(0.5)(a)(30s)² N/A 50 50 m = 0 + (a)(450 s²) 0 N/A 50 m/450 s² = a a = 0.11 m/s² N/A F = ma F = (500kg)(0.11 m/s²) ??? 0.11 N/A F = 55.56 N 30 500

  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. (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) F = ma Δx = vit + ½at² 50 m = (0m/s)(30s)+(0.5)(a)(30s)² N/A 50 50 m = 0 + (a)(450 s²) 0 N/A 50 m/450 s² = a a = 0.11 m/s² N/A F = ma F = (500kg)(0.11 m/s²) ??? 0.11 N/A F = 55.56 N W = Fd 30 W = (55.56 N)(50 m) 500 W = 2778 J

  5. (Wed) You push a 500 kg car that is at rest on a flat road. If the μk 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. (Wed) You push a 500 kg car that is at rest on a flat road. If the μk 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) Fg = mag Fg = (500kg)(9.81m/s²) Fg = 4905 N Fn = Fg Fn = 4905 N N/A 50 Ff = Fnμk Ff = (4905)(0.25) 0 N/A Ff = 1226.25 N N/A 4905 N/A Find Fnet in ‘x’ direction N/A 4905 ??? 0.11 N/A Δx = vit + ½at² 1226 N/A 50m = 0 + (0.5)(a)(30s)² 30 0.11m/s² = a Fnet = ma 1282 N/A 500 Fnet = (500kg)(0.11 m/s²) 55.6 0 Fp = 1282 N Fnet = 55.6 N Fnet = Fp - Ff

  7. (Thu) A 3.00 kg block starts from rest at the top of a 30.0º 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. (Thu) A 3.00 kg block starts from rest at the top of a 30.0º 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) 14.71 N 0 N ???? In the ‘x’ direction….. 5.34 N Fnet = Fg + Fn + Ff Fg,x = magsin(30) 5.34 N = 14.71 N + 0 + Ff Fg,x = (3.00 kg)(9.81 m/s²)(0.5) -9.37 N = Ff Fg,x = 14.71 N

  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. (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 = ½mv² KE = (0.5)(0.150 kg)(35 m/s)² KE = 91.875 kgm²/s² KE = 91.9 J And 1 just ‘cause….

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

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