Monday

December 22, 2014

December 22, 2014

Posted by **FHD** on Saturday, December 7, 2013 at 6:51am.

(a) First, assume the surface is frictionless. What is the magnitude of the final momentum at t1=5s?

P final(t=t1)=

(b) lets now consider a new situation where the object is initially at rest on a rough surface. The coefficient of static friction is 0.2. What is the speed of the block at time t2=5s?

v(t=t2)=

(c) What is the power P provided by the force F(t) at time t3=4s (in Watts) i the case where there is friction (part(b))?

P(t=t3)=

Question No. 2 any ideas.

- Physics Classical Mechanics -
**elli**, Saturday, December 7, 2013 at 3:48pmfor a: 41.67

- Physics Classical Mechanics -
**elli**, Saturday, December 7, 2013 at 4:10pmfor b:

F - f_k = ma

F - f_k = mdv/dt

(F - f_k)dt = mdv

∫(F - f_k)dt = ∫mdv

mv_f - mv_i = (β/3)(t_f)^3 - (f_k)t_f - [(β/3)(t_i)^3 - (f_k)t_i] [Note: v_i = 0 and t_i = 0]

mv_f = (β/3)(t_f)^3 - (f_k)t_f

v_f ≅ 10.83, but that's not the answer, so any suggeston any one???

- Physics Classical Mechanics -
**Daoine**, Saturday, December 7, 2013 at 4:31pmAre you sure about a?

- Physics Classical Mechanics -
**Mets**, Saturday, December 7, 2013 at 4:32pmellie, Thanks. I thought this is the most difficult question. I had trouble with these type of problems throughout the course. I do not have any suggestions. at this point.

- Physics Classical Mechanics -
**Daoine**, Saturday, December 7, 2013 at 4:42pmI did pretty much the same of what you're showing and I got a) 25, b) 2.5 and c) 0.64 I haven't check it yet because it is my last submission and I want to be sure.

- Physics Classical Mechanics -
**Daoine**, Saturday, December 7, 2013 at 4:43pmguys a is 25 for sure

- Physics Classical Mechanics -
**elli**, Saturday, December 7, 2013 at 4:57pmhave you check your data, because for a got a green mark, so A is ok, but b and c were not. There might be some where m = 3 and beta=1.2

- Physics Classical Mechanics -
**Mets**, Saturday, December 7, 2013 at 6:04pmFor question 5 ballistic missile m1= 5*10^24kg; r1=6000km; m2<<m1; alpha=30 degrees; r2=(5/2)r1=15000; G=6.674*10^-11. What is the initial speed of the projectile? My solution:

Vo^2=g*range/sin(theta)

Vo^2=6.674*10^-11*(r1+r2/sin30degrees)

Vo^2=(6.674*10^-11)*(21000/0.5)

Vo^2=0.00000280308

Vo=sqrt(0.00000280308)

Vo=0.00167424

Is this done correctly. Thanks for all your help.

- Physics Classical Mechanics -
**elli**, Saturday, December 7, 2013 at 7:01pmso, no more suggestions for b and c of initial question? and Mets, check the videos from week 13.

- Physics Classical Mechanics -
**Tanero**, Saturday, December 7, 2013 at 9:14pmGuys, please, can you show how you got 25 for a, because I got 41.6 as elli did. For b) my answer is same as ellis too. For c) I used formula for instantaneous power P=FV . So you just need to find values of F and V at t=4 and plug it in there.

- Physics Classical Mechanics -
**Anonymous**, Sunday, December 8, 2013 at 3:22amthere is different m & beta... no wonder if daoine get 25 for a. who has the formula for b & c?

- Physics Classical Mechanics -
**rickross**, Sunday, December 8, 2013 at 3:55amq1 a)

F = dp/dt

dp = Fdt

∫dp = ∫Fdt

p_f - p_i = ∫βt²dt

p_f - 0 = (β/3)(t_f)^3 - (β/3)(t_i)^3

no idea for part b) and c) any help ???

- Physics Classical Mechanics -
**Greco**, Sunday, December 8, 2013 at 7:24amrickross, your solutions for a=41.6 which is comes in contrariety, to a=25.is it surerly right??

- Physics Classical Mechanics -
**rickross**, Sunday, December 8, 2013 at 7:57amthat's is grt formula i got it correct using that formula!

hai Greco help for q8?

- Physics Classical Mechanics -
**Greco**, Sunday, December 8, 2013 at 8:08amrickross still working on it i ll have it soon..

- Physics Classical Mechanics -
**rickross**, Sunday, December 8, 2013 at 9:43amk thanks did u got q1) b ? and c? and if u got it help m ???

- Physics Classical Mechanics -
**Anonymous**, Sunday, December 8, 2013 at 10:40amAnyone can explain the rocket question and the doppler shift question? I got half of each question correct and dont know how to move on.

- Physics Classical Mechanics -
**elli**, Sunday, December 8, 2013 at 12:26pmanswers depend on the values we have

so, for A

F = dp/dt -> dp = Fdt

∫dp = ∫Fdt

p_f - p_i = ∫βt²dt

p_f - 0 = (β/3)(t_f)^3 - (β/3)(t_i)^3

for b:

F - fk = m.a

F - fk = m.dv/dt

(F - fk)dt = m.dv

∫(F - f_k)dt = ∫m.dv

m.vf - m.vi = (β/3)(tf)^3 - (fk)tf - [(β/3)(ti)^3 - (fk)ti] [Note: v_i = 0 and t_i = 0]

m.vf = (β/3)(tf)^3 - (fk)tf

for C) well, I guess we find vf for t=4 as in b, and then

P=W/t -> (1/2m.vf^2)/t

and for m=2 and β=1,

A) 41.67

B) like said before, I got 10.83 but it wasn't the right answer HELP!!!

- Physics Classical Mechanics -
**Just_One**, Sunday, December 8, 2013 at 12:33pmFor part A this is what I did and I got 20 for me and validated 25 for some of you.

F=ma

a=F/m

Because this is not constant acceleration V=at^2. Do not divide by 2! Then P=mv which result in 25 for some of you and 20 for me.

- Physics Classical Mechanics -
**Just_One**, Sunday, December 8, 2013 at 12:53pmLook at this however I am confused about the t^3. I know that is comes from integration but not sure of why the integration if the F(t)Bt^2 is supposed to describe you non linear force in function of a.

answers yah_o_o c_om/

question/

index?qid=20131207034631AA7c7w6

- Physics Classical Mechanics -
**fima**, Sunday, December 8, 2013 at 1:13pmfrom ELLI A its right, but just found;

b) p=m*v, so

v(t)=(integral Ff(t) from (the point where it starts to move) to t) /m

the point where it starts to move:

beta*t^2-m*g*mu=0

t=sqrt(m*g*mu/beta)

v(5)=[beta/3*5^3-m*g*mu*5-beta/3*sqrt(m*g*mu/beta)^3+m*g*mu*sqrt(m*g*mu/beta)]/m

----------------------

c) P=F(4)*v(4)

take F(t)=beta*t^2 and v(4) like in part b)

beta=1 and m=2, uo=0.2

B) vf=5.33

C) P=??

- Physics Classical Mechanics -
**Just_One**, Sunday, December 8, 2013 at 2:35pmI got 6.99 for part B. F=.8 and m=3. Would you please verify. Part A for me is 33.33

Thanks.

- Physics Classical Mechanics -
**Just_One**, Sunday, December 8, 2013 at 2:37pmPower should be the difference in KE when start moving and t final divided by time. Since KE and Power have the same unit and watts is a unit in function of time that is why you need to take the time interval of motion.

- Physics Classical Mechanics -
**fima**, Sunday, December 8, 2013 at 2:41pmI got those new equations from

jiskha . com / display.cgi?id=1386491909

But I'm not sure about those, I'm still waiting for someone to confirm.... any

- Physics Classical Mechanics -
**Anoninho**, Sunday, December 8, 2013 at 5:15pm@Fima

Hello fima... it seems to be the right equation, you see... The force only starts to push the block when it "wins" the Friction Force (Ff). So, its only after 2 seconds has passed. Using your numbers: Ff = m*g*mu = 2*10*0.2 = 4N... F=bt^2 = 1 * (2)^2=4N... So right after 2 seconds, F wins Ff, and the object starts to move on. So, you must integrate from 2 to 5 (question B, and 0 to 4 at C)... You would have:

Fnet = ma

Fnet = mdv/dt

Fnet*dt = mdv, integrating from 2 to 5

Fnet*(t_1-t_0)=mv_1 - mv_0

Fnet*t_1 - Fnet*t_0=mv_1 - mv_0

Fnet = F - Ff and v_0 = 0

(F - Ff)*t_1 - (F - Ff)*t_0 = mv_1

F*t_1 - Ff*t_1 - F*t_0 + Ff*t_0 = mv_1

Changing F =b*t^2, and Ff= m*g*mu

v_1 = [(b/3)*t_1^3 - m*g*mu*t_1 - (b/3)*t_0^3 + m*g*mu*t_0]/m

Putting values in:

v_1 = [(1/3)*5^3 - 2*10*0.2*5 - (1/3)*2^3 + 2*10*0.2*2]/2

v_1 = 27/2 = 13.5 m/s. As far as I can tell. Don't know exactly how you got 5.33, did I do something wrong?

- Physics Classical Mechanics -
**Anoninho**, Sunday, December 8, 2013 at 5:21pm@Fima,

You've said:

B) vf=5.33

But this v_f is only for t=4, that used at question C, at question B is used t=5. Check it out.

- Physics Classical Mechanics -
**fima**, Sunday, December 8, 2013 at 9:09pmThanks, That is v(4) for the next questions, but the procedure is correct

**Answer this Question**

**Related Questions**

Classical Mechanics Physics - A block of mass m= 3 kg is initially at rest on a ...

Classical Mechanics - Need Help Please - A block of mass m= 3 kg is initially at...

mit physyscs - A block of mass 2 kg is initially at rest on a horizontal surface...

physics - A block of mass m= 3 kg is initially at rest on a horizontal surface. ...

Physics - A dockworker applies a constant horizontal force of 79.0 N to a block ...

Physics - A dockworker applies a constant horizontal force of 79.0 N to a block ...

Physics - a 2.5kg block initially at rest on horizontal surface. A horizontal ...

Physics - A dockworker applies a constant horizontal force of 73.0N to a block ...

Physics - A 23.0 kg block is initially at rest on a horizontal surface. A ...

physics - A 20kg block is initially at rest on a rough, horizontal surface. A ...