Originally posted by mtthwThat's correct
OK. This looks like it needs an energy argument (not least because I started off deriving the equation of motion, and it doesn't look too promising). The minimum take off speed required is the one that leaves them stationary at the other side.
Loss of KE = Work done against wind - Loss of PE
mv^2/2 = FD - mgy
where y is the distance dropped
v^2 = 2Fd ...[text shortened]... ils.
Coming back - you haven't provided Tarzan's mass, but the calculation is similar.
A block of wood is sandwiched between two other blocks, and it weighs 95.5 N. If the coefficient of friction between the outer blocks and the sandwiched block is .663, what must be the magnitude of the horizontal compression forces acting on bith sides of the center block to keep it from slipping?
Originally posted by mtthwAhh, true. Missed that part in your solution.
But here you're not swinging from the same height. Unless I'm misinterpreting the question, the finishing level (coming back) is higher than the starting level.
If that's so, the force alone won't be enough to get them back.
I was fiddling around with the energy equation del(KE) + del(PE) = +W, which is the case on the way back, and I came up with a relationship between the mass and the initial velocity:
vi = SQRT(2*g*(hf-hi) - 2*F*d/m)
where:
vi = initial velocity
g = acceleration due to gravity = 9.81 m/s2
hf = final height (i.e. original starting point in part (A)) = -25.7 m
hi = initial height = -35.0 m
F = force of the wind = 110 N
d = horizontal distance = 50 m
m = mass
Below a certain mass, the argument under the square root sign is negative which means that no initial velocity will give you a final velocity of 0 (one of the assumptions made in deriving the above formula). Also, the limit of the initial velocity is SQRT(2*g*(hf-hi) as the mass approaches infinity, which means that if Tarzan and Jane start their swing going at least that fast they are guaranteed to make it to the other side irrespective of Tarzan's weight. Of course, the faster they swing the "hotter" they'll land, so they may want to get the scale out after all. 😉
Originally posted by RamnedDrawing a free body diagram around the block, we can see the forces are as follows:
A block of wood is sandwiched between two other blocks, and it weighs 95.5 N. If the coefficient of friction between the outer blocks and the sandwiched block is .663, what must be the magnitude of the horizontal compression forces acting on bith sides of the center block to keep it from slipping?
sum(Fx) = Fl - Fr
sum(Fy) = -95.5 N + 2*Ff
where:
Fl = force acting on the left side of the block (pointing right)
Fr = force acting on the right side of the block (pointing left)
Ff = force of friction
Solving these equations for the equilibrium conditions sum(Fx) = 0 and sum(Fy) = 0, we get:
Fl = Fr
and
2*Ff = 95.5
Now, since Ff = Fl*mu = Fr*mu, where mu is the coefficient of friction given as 0.663, we get:
2*Fl*mu = 2*Fl*0.663 = 95.5
Fl = 95.5/(2*0.663) = 72.0 N
yes - -
This is hard.
A cannon is attached to a horizontal metal carriage, which can move along the flat ground, but it is connected to a spring, which in turn is attached to a vertical wall of negligible height. The spring is initially stretched with a force constant, k = 2e4 N/m. The cannon fires a 200 kg projectile at a velocity 125 m/s directed at 45 degrees above the horizontal. (A) If the mass of the cannon and metal carriage is 5000 kg, find the recoil speed of the cannon. (B) Determine the maximum extension of the spring. (C) Find the maximum force the spring exerts on the metal carriage. (D) Consider the system as only consisting of the cannon, carriage, and projectile (no spring). Is the momentum conserved during the firing?
Originally posted by RamnedSo the idea is that the spring is in equilibrium at the start?
A cannon is attached to a horizontal metal carriage, which can move along the flat ground, but it is connected to a spring, which in turn is attached to a vertical wall of negligible height. The spring is initially stretched with a force constant, k = 2e4 N/m. The cannon fires a 200 kg projectile at a velocity 125 m/s directed at 45 degrees above the horizont ...[text shortened]... f the cannon, carriage, and projectile (no spring). Is the momentum conserved during the firing?
Mass of cannon/carriage = M
Mass of cannonball = m
Initial speed of cannonball = u
Recoil speed = v
(A) By conservation of momentum horizontally:
MV = mu sin(45)
V = mu/(M sqrt(2))
V = 3.54 m/s
(B) Max extension is x
Final P.E. of spring = Initial K.E. of carriage
kx^2/2 = Mv^2/2
x = v sqrt(M/k)
x = 1.77 m
(C) Maximum force occurs at maximum extension
F = kx
F = 35400 N
(D) Momentum is not conserved, because the system is not isolated. The spring exerts a force on whatever it is tied to, and the carriage pushes down on the ground when it fires the cannonball upwards.
Good! (*Also, the normal force of the ground exerts upward momentum)
(A)Find the acceleration of gravity at the surface of a nuetron star of mass 1.5 solar masses and radius 10 km. (B) Find the weight of a .120 kg baseball on this star. (C) Assume PE = mgh applies, and determine the energy that a 70 kg person would expend climbing a 1.00 cm tall mountain on the star.
Originally posted by Ramned(A) A quick check on Wikipedia tells me that the Sun has a mass of 1.9891 x 10^30 kg, so the neutron star in question would have a mass of 2.98365 x 10^30 kg. Now, using Newton's law of gravitation:
Good! (*Also, the normal force of the ground exerts upward momentum)
(A)Find the acceleration of gravity at the surface of a nuetron star of mass 1.5 solar masses and radius 10 km. (B) Find the weight of a .120 kg baseball on this star. (C) Assume PE = mgh applies, and determine the energy that a 70 kg person would expend climbing a 1.00 cm tall mountain on the star.
F = G*m1*m2/d^2
and dividing through by m2, we get:
a(gravity) = G*m1/d^2
= (6.67 x 10^-11) * (2.98365 x 10^30) / (10 x 10^3)^2
= 1.9901 x 10^12
= 2.0 x 10^12 m/s^2
(B) To find the weight of the baseball at the surface of the neutron star, we sub its mass into Newton's equation:
F = (6.67 x 10^-11) * (2.98365 x 10^30) * (0.120) / (10 x 10^3)^2
= 2.3881 x 10^11
= 2.4 x 10^11 N
(C) Using the given equation, and assuming gravity is constant over the distance, the minimum amount of energy needed to move a 70 kg person from the surface to the top of a 1.00 cm "mountain" is as follows:
E = mgh(final) - mgh(initial)
= mg*(h(final) - h(initial))
= (70) * (2.0 x 10^12 m/s^2) * (0.0100 - 0)
= 1.3931 x 10^12
= 1.4 x 10^12 J
heh heh: Here's how 😉
Use Kepler's 3rd Law:
K = ((4)(pi^2)/(G*Mass of the Sun))
Where K is a constant, 2.97e-19 and G is 6.667e-11. You can then solve for the Mass of the sun. You can use this same equation to find the mass of the earth and any other planet:
K = ((4)(pi^2)/(G*Mass 'planet'😉) * (r^3)
where r is the distance from the center of the planet to the center of the sun.
Due to the gravitational torque exerted by the moon on the Earth, our planet's period of roatation slows at a rate on the order of 1ms/ century. (A) Determine the order of magnitude of Earth's angular acceleration. (B) Find the order of magnitude of the torque. (C) Find the order of magnitude if the size of a wrench an ordinary person would need to exert the torque found in (B). Assume the person can brace his feet against a solid firmament.
(The answers will be approximate.)
Originally posted by Ramneda = (w2-w1)/t
Due to the gravitational torque exerted by the moon on the Earth, our planet's period of roatation slows at a rate on the order of 1ms/ century. (A) Determine the order of magnitude of Earth's angular acceleration. (B) Find the order of magnitude of the torque. (C) Find the order of magnitude if the size of a wrench an ordinary person would need to exert the ...[text shortened]... the person can brace his feet against a solid firmament.
(The answers will be approximate.)
w1 = 2pi/(24*3600)
= 7.272205217 x 10^-5 rad/s
w2 = 2pi/(24*3600 + 0.001)
= 7.272205132 x 10^-5 rad/s
t = 100*365.25*24*3600
= 3.15576 x 10^9 s
a = (-8.5 x 10^-13)/3.15576 x 10^9
A: a = -2.7 x 10^-22 rad/s^2
T = Ia
I = 0.4MR^2 (assuming a spherical earth)
= 0.4*(5.97 × 10^24)*(6.4 x 10^6)^2
= 9.78 x 10^37
B: T = 2.64 x 10^16 Nm
I'm going to guess a normal person could exert a force of 1kN.
Fd = T
1000d = 2.64 x 10^16
d = 2.64 x 10^13 m
C: d = 26,400,000,000 km