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
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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?