(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)

Δx = vit + ½at²

(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 s²)50 m/450 s² = aa = 0.11 m/s²

50 m = (0m/s)(30s)+(0.5)(a)(30s)²

x y

Δxvi

vf

a

t

m

N/A

N/A

N/A

N/A

50

30

???

0

500

F = ma

0.11 F = ma F = (500kg)(0.11 m/s²)

(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)

Δx = vit + ½at²

(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 s²)50 m/450 s² = aa = 0.11 m/s²

50 m = (0m/s)(30s)+(0.5)(a)(30s)²

x y

Δxvi

vf

a

t

m

N/A

N/A

N/A

N/A

50

30

???

0

500

F = ma

0.11 F = ma F = (500kg)(0.11 m/s²)W =

FdF = 55.56 NW = (55.56 N)(50 m)

(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)

50m = 0 + (0.5)(a)(30s)²

Fg = mag Fg = (500kg)(9.81m/s²)

(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)

Fp = 1282 N

Fn = Fg

Fn = 4905 NFg = 4905 N

x y

Δxvi

vf

a

t

m

N/A

N/A

N/A

N/A

50

30

???

0

500

0.11

Ff = Fnμk Ff = (4905)(0.25)

Find Fnet in ‘x’ directionFf = 1226.25 NΔx = vit + ½at²

x y

Fg

Fn

Ff

Fp

Fnet

N/A

N/A

N/A

N/A

0

4905

4905

1226

0.11m/s² = a

Fnet = maFnet = (500kg)(0.11 m/s²)Fnet = 55.6 N

55.6

1282

Fnet = Fp - Ff

(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)

(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)

x Dir y Dir

Fg

Fn

Ff

Fnet

14.71 N

0 N

5.34 N

Fg,x = magsin(30)

Fg,x = (3.00 kg)(9.81 m/s²)(0.5)

Fg,x = 14.71 N

Fnet = Fg + Fn + Ff

5.34 N = 14.71 N + 0 + Ff

-9.37 N = Ff

????

In the ‘x’ direction…..

(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 = 91.9 J

KE = ½mv²

(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)²KE = 91.875 kgm²/s²

And 1 just ‘cause….

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