A block of mass m= 3 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)=βt2, where β= 1.2 N/s2. We stop pushing at time t1=5 s [F(t)=0 for t>t1].
(a) First, assume the surface is frictionless. What is the magnitude of the final momentum of the block at t1=5 s? (in kg m/s)
pfin(t=t1)=
(b) Let us now consider a new situation where the object is initially at rest on a rough surface. The coefficient of static friction is μs=0.2. What is the speed of the block at time t2=5 s?. For simplicity, we take static and kinetic friction coefficients to be the same, μs=μk and consider g=10 m/s2.
v(t=t2)=
(c) What is the power P provided by the force F(t) at time t3=4 s (in Watts) in the case where there is friction (part (b)) ?
P(t=t3)=
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I found a). = 50 by using the formula Beta/3*t^3 But am unsure about the others. Help Please.
F = m dV/dt
m dV = F dt
so
change in momentum = integral of F dt
F = 1.2 t^2
integral F dt = 1.2 t^3/3
= 50 at t = 5 yes
for part b
F = 1.2 t^2 - 10(3)(.2) = 1.2t^2-6
integral F dt = 1.2 t^2/3 - 6 t
for part 3
P = F dx/dt
F = 1.2 t^2 - 6
at 4 sec F = 13.2 Newtons
to get dx/dt at t = 4 we could use energy methods or brute force
brute force:
d^2x/dt^2 = F/m = .4 t^2 - 2
dx/dt = (.4/2)t^3 - 2 t
at t = 4
dx/dt = 12.8 - 8 = 4.8 m/s
so
power at t = 4 is 13.2 * 4.8
= 63.4 Watts
>to get dx/dt at t = 4 we could use energy methods >or brute force
>brute force:
>d^2x/dt^2 = F/m = .4 t^2 - 2
>dx/dt = (.4/2)t^3 - 2 t
where does the 1/2 come from in (.4/2)t^3 ?
sorry, .4/3
hai damon this integral is correct for part b
F = 1.2 t^2 - 10(3)(.2) = 1.2t^2-6
integral F dt = 1.2 t^2/3 - 6 t ????
F(t)=beta*t^2
Ff(t)=beta*t^2-m*g*mu
a) p(t)=integral F(t) from 0 to t
p(t1)=beta/3*t1^3
---------------------------
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)
HEY BW!!
following your procedure, did you check your answers?
hai BW I USED the above formula and i got b and c wrong i got negative value plz can u help m ???? this is my question 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)=βt2, where β= 0.8 N/s2. We stop pushing at time t1=5 s [F(t)=0 for t>t1].
(a) First, assume the surface is frictionless. What is the magnitude of the final momentum of the block at t1=5 s? (in kg m/s)
pfin(t=t1)=
correct
(b) Let us now consider a new situation where the object is initially at rest on a rough surface. The coefficient of static friction is μs=0.2. What is the speed of the block at time t2=5 s?. For simplicity, we take static and kinetic friction coefficients to be the same, μs=μk and consider g=10 m/s2.
v(t=t2)=
incorrect
(c) What is the power P provided by the force F(t) at time t3=4 s (in Watts) in the case where there is friction (part (b)) ?
P(t=t3)=
@fem: yes, i would not post it if I had not checked it.
@kumar: probably you integrated from 0 to 5?
F(t)=0.8*t^2
Ff(t)=0.8*t^2-4
b) time where it starts to move:
0.8*t^2-4=0
->t=sqrt(4/0.8)=sqrt(5)
v(5)=[0.8/3*t^3-4t] integral from sqrt(5) to 5 /2
=9.65 (rounded)
c)v(4)=3.51 (rounded)
F(4)=0.8*4^2=12.8
P(4)=44.99 (rounded)
ok, so considering b=1 and m=2
vf t=5= 5.33
P= F(4)*v(4)=85.28
WEll, thanks @BW, by the way, have you donde the ruler problem?
no, that's the only one I'm not done with, I'll try tomorrow..
and the question for the merry?
I'm missing the ruler and the merry.
merry is just a little conservation of momentum..
m1: mass merry-go-round
m2: mass sledgehammer
v1: velocity merry-go-round before collision (=0)
v2: velocity sledgehammer before collision
v1': velocity merry-go-round after collision
v2': velocity sledgehammer after collision (=0)
(m1*v1)+(m2*v2)=(m1*v1')+(m2*v2')
m2*v2=m1*v1'
v1'=m2*v2/m1
energy difference:
((0.5*m1*v1^2)+(0.5*m2*v2^2))-((0.5*m1*v1'^2)+(0.5*m2*v2'^2))
|deltaE|=0.5*m2*v2^2-0.5*m1*v1'^2
that was really easy, thanks!!!
any hint for the ruler, so I can try it out??
I have not yet tried to solve it, maybe you'll find something useful in the thread with id=1386456397 here on jiskha
@fima,
Have you submited that deltaE value? I thought it had to take in account Kinect Rotational Energy of the Merry-go-round... But the energy loss is about the same (974.234 without it, 951.129 considering it.) Just wanting to confirm.
bw,
For the above question (the block of mass 3kg one) I am still getting the incorrect answer for your answer of c.) 44.99
Is it possible this is incorrect still?
thanks bw your formula worked for both q1 and merry, thanks so much for sharing with us :)
@Jeff: 44.99 is the solution for kumars values, yours are:
c)
Ff(t)=1.2*t^2-6
->t=sqrt(6/1.2)=sqrt(5)
v(4)=3.51 (rounded)
F(4)=1.2*4^2=19.2
P(4)=67.48 (rounded)
You beauty, Thanks bw.
All I have left is the falling ruler if you have any advice for that one. Otherwise thanks a million!
please tell the formula for b & c,
my beta is 0.6 and mass is 2kg
jeff can u say the answers for this questions
please tell the formula for b & c,
my beta is 1.2 and mass is 3kg
Please help me, my beta is 0.8 and mass is 3 kg, I got only the a) part correct, b) and c) are wrong :(
woooooooooo thx @Hw !did u got this correct
Consider a rocket in space that ejects burned fuel at a speed of vex= 2.0 km/s with respect to the rocket. The rocket burns 11 % of its mass in 250 s (assume the burn rate is constant).
(a) What is the speed v of the rocket after a burn time of 125.0 s? (suppose that the rocket starts at rest; and enter your answer in m/s)
v=
incorrect
(b) What is the instantaneous acceleration a of the rocket at time 125.0 s after the start of the engines?(in m/s2)
a=
help?