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## Homework Statement

Two pucks (5 kg each) made of Teflon are on a long table, also made of Teflon. Puck A is sitting at

rest on the left end of the table. Puck B is 15 m away at the right hand end of the table, and is

travelling toward Puck A with an initial speed of 0.5 m/s. A person on the left waits 1.5 seconds and

then pushes Puck A forward from rest with a force of 20 N applied at an angle of 30° to the horizontal.

If that force is maintained at a steady rate for the entire question, how far will puck A travel before it

strikes puck B? (The coefficient of friction between the Teflon and Teflon is 0.04.

**2. The attempt at a solution:**

Fax = 20NSin30

Fax = 10N

Fay = 20NCos30

Fay = 17.32N

Fg = mg

Fg = (5kg)(9.8m/s^2)

Fg = 49N

__Puck A__Fax = 20NSin30

Fax = 10N

Fay = 20NCos30

Fay = 17.32N

Fg = mg

Fg = (5kg)(9.8m/s^2)

Fg = 49N

**Fn = Fay + Fg**

Fn =17.32N + 49N

Fn = 66.32

Ff = uFn

Ff = (0.04)(66.32N)

Ff = 2.65

Fnet = Fax - Ff

Fnet = 10N - 2.65

Fnet = 7.35N

a = Fnet/m

a = 7.35N/5kg

a = 1.47m/s^2

d = v1t + 1/2at^2

15-x = 1/2(1.47)(y-1.5)^24

x = -0.74y^2 + 2.22y +13.33

Fg = mg

Fg = (5kg)(9.8m/s^2)

Fg = 49N

Ff = uFn

Ff = (0.04)(49N)

Ff = 1.96N

Fn =17.32N + 49N

Fn = 66.32

Ff = uFn

Ff = (0.04)(66.32N)

Ff = 2.65

Fnet = Fax - Ff

Fnet = 10N - 2.65

Fnet = 7.35N

a = Fnet/m

a = 7.35N/5kg

a = 1.47m/s^2

d = v1t + 1/2at^2

15-x = 1/2(1.47)(y-1.5)^24

x = -0.74y^2 + 2.22y +13.33

__Puck B__Fg = mg

Fg = (5kg)(9.8m/s^2)

Fg = 49N

Ff = uFn

Ff = (0.04)(49N)

Ff = 1.96N

**Since there is no applied force Fnet = -Ff**

a = Fnet/m

a = -1.96/5kg

a = -0.39m/s^2

a = Fnet/m

a = -1.96/5kg

a = -0.39m/s^2

**d = v1t + 1/2at^2****x = (0.5)(y) + 1/2(-0.39)(y)^2**

**-0.74y^2 + 2.22y +13.33 = 0.5 - 0.195y^2****0 = 0.545y^2 - 1.72y-13.33****I then put it in the quadratic formula and got:**

y = 6.77s and y = -3.61s(omited)

y = 6.77s and y = -3.61s(omited)

d = (0m/s)(6.77s-1.5s) + 1/2(1.47m/s^2)(6.77-1.5s)^2

d = 20.41m(This answer doesn't make sense)

__Puck A Distance__**d = v1t + 1/2at^2**d = (0m/s)(6.77s-1.5s) + 1/2(1.47m/s^2)(6.77-1.5s)^2

d = 20.41m(This answer doesn't make sense)

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