A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. A horizontal force of 2.5 N is applied o
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Answer:
(a) v-bullet = 399.04 m/s
(b) I = 2.38 kg m/s
(c) T = 2.59 N
Explanation:
(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.
The potential energy is given by:
(1)
U: potential energy
M: mass of the wood block = 1 kg
m: mass of the bullet = 6g = 6.0*10^-3 kg
g: gravitational constant = 9.8m/s^2
h: distance to the ground
The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:
The potential energy is:
Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:
Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:
(2)
v1: initial velocity of the wood block = 0m/s
v2: initial speed of the bullet
v: speed of bullet and block = 2.38m/s
You solve the equation (2) for v2:
The speed of the bullet before the impact with the wood block is 399.04 m/s
(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:
The impulse is 2.38 kgm/s
(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:
The force on the cord after the impact is 2.59N
Answer:
Δt =
Explanation:
Hi there!
Using the equation of speed for the whole trip, we can obtain the time each one needed to cover the distance D.
The speed (v) is calculated by dividing the traveled distance (d) over the time needed to cover that distance (t):
v = d/t
Rick traveled half of the distance at Vr and the other half at Vw. Then, when v = Vr, the distance traveled was D/2 and the time is unknown, Δt1:
Vr = D/ (2 · Δt1)
For the other half of the trip the expression of velocity will be:
Vw = D/(2 · Δt2)
The total time traveled is the sum of both Δt:
Δt(total) = Δt1 + Δt2
Then, solving the first equation for Δt1:
Vr = D/ (2 · Δt1)
Δt1 = D/(2 · Vr)
In the same way for the second equation:
Δt2 = D/(2 · Vw)
Δt + Δt2 = D/(2 · Vr) + D/(2 · Vw)
Δt(total) = D/2 · (1/Vr + 1/Vw)
The time needed by Rick to complete the trip was:
Δt(total) = D/2 · (1/Vr + 1/Vw)
Now let´s calculate the time it took Tim to do the trip:
Tim walks half of the time, then his speed could be expressed as follows:
Vw = 2d1/Δt Where d1 is the traveled distance.
Solving for d1:
Vw · Δt/2 = d1
He then ran half of the time:
Vr = 2d2/Δt
Solving for d2:
Vr · Δt/2 = d2
Since d1 + d2 = D, then:
Vw · Δt/2 + Vr · Δt/2 = D
Solving for Δt:
Δt (Vw/2 + Vr/2) = D
Δt = D / (Vw/2 + Vr/2)
Δt = D/ ((Vw + Vr)/2)
Δt = 2D / (Vw + Vr)
The time needed by Tim to complete the trip was:
Δt = 2D / (Vw + Vr)
Let´s find the diference between the time done by Tim and the one done by Rick:
Δt(tim) - Δt(rick)
2D / (Vw + Vr) - (D/2 · (1/Vr + 1/Vw))
= Δt
Let´s check the result. If Vr = Vw:
Δt = 2D/2Vr - D/2Vr - D/2Vr
Δt = D/Vr - D/Vr = 0
This makes sense because if both move with the same velocity all the time both will do the trip in the same time.