1. Two blocks are connected by a light string that passes over two frictionless

Question

1. Two blocks are connected by a light string that passes over two frictionless pulleys as in the figure below. The block of mass m2 is attached to a spring of force constant k and m1 > m2. If the system is released from rest, and the spring is initially not stretched or compressed, find an expression for the maximum displacement d of m2. (Use any variable or symbol stated above along with the following as necessary: g.)

d =

2. A daredevil on a motorcycle leaves the end of a ramp with a speed of 34.0 m/s as in the figure below. If his speed is 32.5 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance.
in m

(a) What is the initial total mechanical energy of the projectile?
in J

(b) Suppose the projectile is traveling 102.1 m/s at its maximum height of y = 360 m. How much work has been done on the projectile by air friction?
in J

(c) What is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up?
in m/s

Explanation / Answer

1)

First define your initial and final position. To make it easier, the initial position when the system is at rest. We will define y=0 at the position of the masses at rest. When the system is released, m1 drops by a distance d, m2 rises by the same distance d, and the spring is stretched by the same distance d as well, since they are all connected. Initially the system has no kinetic energy or potential energy. After the system is displaced a distance, d, after release the system has gained potential energy - both gravitational potential energy, PEg and Spring Potential energy, PEs. Thus,
E(initial) = E (final)
0 = PEg (for m1) + PEg (for m2) + PEs
0 = m1g(-d) + m2gd + 0.5kd^2 (FY1: d is negative for m1 because m1 being drops down)
Now, divide both sides by d
0 = -m1g + m2g + 0.5kd
Isolate for d
2g (m1 - m2) / k = d

2)

He is traveling upwards; but his acceleration is downwards (gravity). The equation to use is v^2=u^2+2aS. We want to find out the distance he reaches (i.e. th maximum height), which is 'S' in the equation. We are given (or know) the other variables:
u= 34.0m/s
v=32.5m/s
a=-9.8m/s^2

So we isolate and solve for S:
S=(V^2 - u^2)/2a
=(32.5^2 - 34.0^2)/2(-9.8)
= 5.08 m

3)

  (a) the initial total mechanical energy of the projectile
= 0.5 *42 * 144^2
= 435456 J --answer

(b) Suppose the projectile is traveling 87.9 m/s at its maximum height of y =360 m.
height travelled = 360 - 144 = 216 m
PE = 42 * 9.81 *216 = 88997 J
KE = 0.5 * 42 * 102.1 ^2 = 218912.61 J

work has been done on the projectile by air friction
= 435456 -88997 - 218912.61
= 127546.39 J ----answer

(c) if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up
work done by friction = 1.5 * 127546.39 J = 191319.585 J

PE lost = 42 * 9.81 * 360= 148327.2 J
KE gain = 0.5 * 42 * v^2

435456 = 191319.585 + 148327.2 + 0.5 * 42 * v^2
v =309 m/s
answer

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