Physics Classical Mechanics
posted by FHD on .
A block of mass m=2 kg is initially at rest on a horizontal surface. At time t=0, we begin pushing on it with a horizontal force that varies with time as F(t)=Beta t^2, where Beta = 1.0 N/s^2. We stop pushing at time t1, 5s [F(t)=0 for t>1].
(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.

for a: 41.67

for 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??? 
Are you sure about a?

ellie, 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.

I 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.

guys a is 25 for sure

have 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

For 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. 
so, no more suggestions for b and c of initial question? and Mets, check the videos from week 13.

Guys, 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.

there is different m & beta... no wonder if daoine get 25 for a. who has the formula for b & c?

q1 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 ??? 
rickross, your solutions for a=41.6 which is comes in contrariety, to a=25.is it surerly right??

that's is grt formula i got it correct using that formula!
hai Greco help for q8? 
rickross still working on it i ll have it soon..

k thanks did u got q1) b ? and c? and if u got it help m ???

Anyone can explain the rocket question and the doppler shift question? I got half of each question correct and don't know how to move on.

answers 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!!! 
For 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. 
Look 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 
from 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^2m*g*mu=0
t=sqrt(m*g*mu/beta)
v(5)=[beta/3*5^3m*g*mu*5beta/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=?? 
I got 6.99 for part B. F=.8 and m=3. Would you please verify. Part A for me is 33.33
Thanks. 
Power 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.

I 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 
@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_1t_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? 
@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. 
Thanks, That is v(4) for the next questions, but the procedure is correct