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HawKMX2004
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->Forces with Friction help<- ASAP please
I have two problems that i didnt quite understand, i would appreciate any help i could get, my work follows the questions
1.) A block with a mass of .2 kg is slid across a patch of ice with an initial velocity of 20 m/s. After Traveling 70 m, the velocity is 10 m/s. Find the following:
a.) The acceleration of the puck a = acceleration
Vf^2=Vi^2 + 2a(Xf-Xi)
Vf^2 / a = Vi^2 +2(Xf-Xi)
10^2 / a = 20^2 +2(70-0)
100 / a = 400 + 140
100 / a = 540
a = 100/540
a = .19 m/s/s
b.) The Force of Friction - Ff = Force of Friction
Ff = ma
Ff = .2 kg(.19 m/s/s)
Ff = .04N
c.) Coefficient of Friction - U = Coefficient of Friction, FN = Force Normal
U = Ff / FN FN = mg = .2 (9.8) = 1.96N
U = .04 / 1.96
U = .02
2.) A 40 kg box is being pulled by a rope to the right with a tension of (T = 300 N) which is at a 37 degree angle to the horizontal. Another force of 50N is applied to the box pulling left perfectly horizontally. The coefficient of friction between the box and the floor is .4 The box is moving.
a.) Find the Force with which the floor pushes up on the box
Force Gravity = Force Pushing Up
F = ma
F = 40kg(9.8m/s/s)
F = 392N = 390N with sigfigs
b.) Find the acceleration of the box - Fr = Resisting force
Tx = cos(Theta)T Ty = sin(Theta)T
Tx = cos(37)(300) Ty = sin(37)(300)
Tx = 240N Ty = 180.5N
Fnet = Tx - 50N - Ff = ma
Fnet = 240N - 50N - Ff = ma
Fnet = 190N - FnUk = ma
Fnet = 190N - 210N(.4) = ma
Fnet = 190N - 84N = ma
Fnet = 106N / m = a
Fnet = 106 / 40 = a
Fnet = 2.65 m/s/s = a
Any help i could get on either of those two problems would be greatly appreciated...if i got them right, please let me know, because i think I'm starting to get it. Thank You
I have two problems that i didnt quite understand, i would appreciate any help i could get, my work follows the questions
1.) A block with a mass of .2 kg is slid across a patch of ice with an initial velocity of 20 m/s. After Traveling 70 m, the velocity is 10 m/s. Find the following:
a.) The acceleration of the puck a = acceleration
Vf^2=Vi^2 + 2a(Xf-Xi)
Vf^2 / a = Vi^2 +2(Xf-Xi)
10^2 / a = 20^2 +2(70-0)
100 / a = 400 + 140
100 / a = 540
a = 100/540
a = .19 m/s/s
b.) The Force of Friction - Ff = Force of Friction
Ff = ma
Ff = .2 kg(.19 m/s/s)
Ff = .04N
c.) Coefficient of Friction - U = Coefficient of Friction, FN = Force Normal
U = Ff / FN FN = mg = .2 (9.8) = 1.96N
U = .04 / 1.96
U = .02
2.) A 40 kg box is being pulled by a rope to the right with a tension of (T = 300 N) which is at a 37 degree angle to the horizontal. Another force of 50N is applied to the box pulling left perfectly horizontally. The coefficient of friction between the box and the floor is .4 The box is moving.
a.) Find the Force with which the floor pushes up on the box
Force Gravity = Force Pushing Up
F = ma
F = 40kg(9.8m/s/s)
F = 392N = 390N with sigfigs
b.) Find the acceleration of the box - Fr = Resisting force
Tx = cos(Theta)T Ty = sin(Theta)T
Tx = cos(37)(300) Ty = sin(37)(300)
Tx = 240N Ty = 180.5N
Fnet = Tx - 50N - Ff = ma
Fnet = 240N - 50N - Ff = ma
Fnet = 190N - FnUk = ma
Fnet = 190N - 210N(.4) = ma
Fnet = 190N - 84N = ma
Fnet = 106N / m = a
Fnet = 106 / 40 = a
Fnet = 2.65 m/s/s = a
Any help i could get on either of those two problems would be greatly appreciated...if i got them right, please let me know, because i think I'm starting to get it. Thank You
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