PHY131Ch 7 – 9 ExamName

Your car impacts with a dense foam wall. The force exerted on your car varies in time by the given equation of –(340+ 80t3). If your 2000 kg car comes to a stop in 5 seconds, what was the initial velocity? / F dt = Δ m v
-(340 + 80t3)dt = 2000 (0 – v0)
-340t – 20 t4 = -2000 v0
340(5) + 20(54) = 2000 v0
v0 = 7.1 m/s
Three forces are applied to a 4.00 kg trunk that moves leftward by 3.00 m over a floor with a coefficient of friction of 0.3. The force magnitudes are F1 = 6.00 N, F2 = 5.00 N, and F3 = 8.00 N, and the indicated angle is θ = 60°. (a) During the displacement, what is the net work done on the trunk? (b) Does the kinetic energy of the trunk increase or decrease? /
Work = Fapplied Δx
Work = (6 – (5)cos60° - Ff) Δx
Work = (3.5 - μ FN ) Δx
Work = (3.5 - .3 (8 - (5)sin60°)3m= 7.2 Nm / Work = 7.2 Nm = Δ( ½ m v2)
v = 1.90 m/s, thus KE increases

A 30 gram arrow, buries itself 10 cm into a stationary 300 gram foam target. This foam target is now placed on a frictionless horizontal surface, and a second 30 g arrow is shot into the target. To what depth will the 2nd arrow penetrate the form target?

1st Case
Wfoam = K
F d = Kf– Ki
F(.1)= 0 - ½(.03)v2
F = -0.15 v2 / 2nd Case
mv = (M+m+m)vf
0.03 v = (M+0.06) vf
vf = 0.03 v /(.36) / K0 – Ffoam (d)=Kf
½ 0.03 v2- F d= ½ 0.36 vf2
½ 0.03 v2- (-.15 v2)d = ½ 0.36 (0.03v /(.36))2
v2 cancels out
d = 0.0916 m or 9.16 cm

PHY131Ch 7 – 9 ExamName


Block R
K0 – Ff d = Kf
½ m vo2 – μR mg (dx) = 0
½ m vo2 – (.1+ ½ x)mg (dx) = 0
½ m vo2 – mg ∫(.1+ ½ x)(dx) = 0 / A stationary block explodes into two pieces L and R that slide across a frictionless floor and then into regions with friction, where they stop. Piece L, with a mass of 2.0 kg, encounters a coefficient of kinetic friction μL = 0.50 and slides to a stop in distance xL = 0.40 m. Piece R encounters a coefficient of kinetic friction μR = (0.1 + .5x) and slides to a stop in distance xR = 0.80 m. What was the mass of the block?
½ m vo2 – mg (.1x + x2) (from 0 to .8) = 0
½ m vo2 – .08mg + .64mg = 0
½ m vo2 = .72mg
vR = 3.8 m/s
Block L
K0 – Ff d = Kf
½ 2 vo2 – μLmg (0.4) = 0
vL = 2 m/s / po = 0 kgm2 = pf
so final momentum to the left = to the right
mL vL = mR vR
2(2) = mR3.8
mR = 1.05 kg
mT = 2 + 1.05 kg = 3.05 kg
A 800 kg elevator starts from rest. It moves upward for 3 s with constant acceleration until it reaches its cruising speed of 2.00 m/s. (a) What is the average power of the elevator motor during this period? (b) How does this power compare with its power when it moves at its cruising speed / P = Work / Δt
P = Fw Δy / Δt
P = Fw v t / Δt
P = F v
a = Δv / Δt
a = 2 / 3 s
a = 0.667 m/s2 / d = ½ at2
d = 3 m / Work = Fapplied * Δy
Work = 800(g+a) 3
Work = 26000 Nm / P = Work / time
P = 26000 J / 3s
P = 8670 Watts / (b) P = Fw vy
P = 8000(2)
P = 16000 W
A 3.0 kg breadbox on a frictionless incline of angle θ = 30˚ is connected, by a cord that runs over a massless and frictionless pulley, to a light spring of spring constant k = 200 N/m. The box is released from rest when the spring is unstretched. (a) What is the speed of the box when it has moved 10.0 cm down the incline? And what is the (b) magnitude of the box's acceleration at the instant the box momentarily stops?
No non-conservative forces, thus TE is conserved. /
TEtop = Ug + Uspring + KE
TEtop = mg h + 0 + 0
TEtop = 3g(sin30(0.1m))
TEtop = 1.5 Nm
TE10cm = Ug + Uspring + KE
1.5 = 0 + ½200(.1)2 + ½3v2
v = 0.577 m/s / TEtop = Ug + Uspring + KE
TEtop = mg h + 0 + 0
TEtop = 30 sin30*d
TEtop = 15d
TEbottom = Ug + Uspring + KE
TEbottom = 0 + ½200(d)2 + 0 / TEtop = TEbottom
15d = ½200(d)2
d = 0.15 meters
F = -kx = ma
F = 200(.15) = 3(a)
a = 10 m/s2