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3 years ago

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A 56 kg pole vaulter falls from rest from a height of 5.1 m onto a foam rubber pad. The pole vaulter comes to rest 0.29 s after

landing on the pad.
Required:
a. Calculate the athlete's velocity just before reaching the pad
b. Calculate the constant force exerted on the pole vaulter due to the collision

1 answer:

never [62]3 years ago

6 0

Answer:

b. calculate the constant force exerted on the pole vaulter due to the collision

mbgiclbxhkr and you can see the pole and you are not to be a great friend of the day and night sweets is my first day of your life and my friends is my friend

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Please help me with part b.

Lerok [7]

I'm sorry but I don't see anything that says part B
.

8 0

3 years ago

A cord of negligible mass runs around two massless, frictionless pulleys. A canister with mass m = 20 kg hangs from one pulley.

photoshop1234 [79]

(a) 196 N

The equation of the forces on the side of the cord where the force F is applied is:

$F-T=0$ (1)

where T is the tension in the cord.

On the other side of the cord, the equation of the forces on the canister is

$T-mg = ma$ (2)

where

m = 20 kg is the mass of the canister

$g=9.8 m/s^2$ is the acceleration of gravity

a is the acceleration

From (1),

$T=F$

Substituting into (2),

$F-mg = ma\\F=m(g+a)$

We want the canister to move at constant speed, so

a = 0

And therefore:

$F=mg=(20)(9.8)=196 N$

b) 2.0 cm

The cord is inextensible, this means that the acceleration of its parts are the same. Therefore, the acceleration of the free end must be the same as the acceleration of the canister: and this means that the two parts also cover the same distance in the same time.

Therefore, the free end of the cord must be moved exactly the same as the canister, by 2.0 cm.

c) 3.92 J, the same

The work done by the tension in the cord is

$W_T = T d$

where

T is the tension

d = 2.0 cm = 0.02 m is the displacement

As we said in part (a), the tension in the cord is equal to the force applied to the free end:

T = F

So

T = 196 N

Therefore, the work done by the tension is

$W=(196)(0.02)=3.92 J$

And since the force applied (F) is the same, then the work done by you when pulling the cord is exactly the same.

(d) -3.92 J

The weight of the canister is

$F_g = mg =(20 kg)(9.8 m/s^2)=196 N$

However, the direction of the force of gravity is opposite to the displacement. Therefore, the work done by gravity is negative:

$W_g = - F_g d$

And substituting,

$W_g=-(196)(0.02)=-3.92 J$

(e) Zero

The net work done on the canister can be simply calculated by adding the work done by the tension in the cord and the weight of the canister:

$W=W_T+W_g = 3.92 + (-3.92 ) = 0$

This is in agreement with the work-energy theorem, which states that the work done on an object is equal to its change in kinetic energy. In this situation, the canister is moving at constant speed, so its kinetic energy is not change: therefore,

$\Delta K = 0$ (change in kinetic energy = 0)

and so, the work done on it is also zero.

(f) The pulley system changes the direction of the force applied

This is a simple pulley system, which means that the system does not multiply the force applied in input. In fact, the mechanical advantage of the system is

$MA=\frac{F_{out}}{F_{in}}$

where:

$F_{out}$ is the output force, which is the weight of the canister

$F_{in}$ is the force in input, which is F

So, the mechanical advantage is 1:

$MA=\frac{196 N}{196 N}=1$

From a point of view of energy, therefore, there is no advantage in this system.

However, the advantage offered by the pulley system concerns the direction of the force: in fact, it changes the direction of the applied force (which is F, downward) into the tension of the cord (which is upward on the canister).

6 0

3 years ago

A 70 kg human sprinter can accelerate from rest to 10 m/s in 3.0 s. During the same time interval, a 30 kg greyhound can go from

ladessa [460]

Answer:

$P_1 = 1166.7 Watt$

$P_2 = 2000 Watt$

Explanation:

Average power for the human sprinter is given as

$Power = \frac{\Delta E}{\Delta t}$

so we have

$P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}$

$P = \frac{\frac{1}{2}(70)(10^2) - 0}{3}$

$P_1 = 1166.7 Watt$

Average power for greyhound is given as

$P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}$

$P = \frac{\frac{1}{2}(30)(20^2) - 0}{3}$

$P_2 = 2000 Watt$

3 0

3 years ago

What is energy converted to if it is not used to do work?

Romashka [77]

Answer:

the human body isn't very efficient at converting food into useful work. The human body is less than 5% efficient most of the time. The rest of the energy is converted to heat, which may or may not be useful, depending on how cool or warm a person wants to be.

Explanation:

3 0

2 years ago

A negatively charged particle is moving to the right, directly above a wire have a current flowing to the right. In which direct

Varvara68 [4.7K]

Answer:

C) upward

Explanation:

The problem can be solved by using the right-hand rule.

First of all, we notice at the location of the negatively charged particle (above the wire), the magnetic field produced by the wire points out of the page (because the current is to the right, so by using the right hand, putting the thumb to the right (as the current) and wrapping the other fingers around it, we see that the direction of the field above the wire is out of the page).

Now we can apply the right hand rule to the charged particle:

- index finger: velocity of the particle, to the right

- middle finger: direction of the magnetic field, out of the page

- thumb: direction of the force, downward --> however, the charge is negative, so we must reverse the direction --> upward

Therefore, the direction of the magnetic force is upward.

3 0

3 years ago