What is the kinetic energy f a 25kg object movingat a velocity of 10m/s?

A 465 g block slides along a frictionless surface at a speed of 0.35 m/s. It runs into a horizontal massless spring with spring

katovenus [111]

Answer:

a) The duration, during which the block remain in contact with the spring is 0.29 s

b) The period of the simple harmonic oscillatory motion depends only on the mass and spring constant, therefore when the speed is doubled, the duration of contact remains the same as 0.29 s.

Explanation:

Mass of the block = 465 g

Surface speed = 0.35 m/s

Spring constant , k = 54 N/m

$T = 2\times \pi \times \sqrt{\frac{m}{K} } = 2\times \pi \times \sqrt{\frac{0.465}{54} }$ = 0.58 s

a) Since the period for the oscillatory motion is 0.58 s, then the time when the block and spring remain in contact is T/2 = 0.29 s

b) When the speed is doubled, we have

$T = 2\times \pi \times \sqrt{\frac{m}{K} }$

Therefore, since T is only dependent on the mass, m and the spring constant, K, then the time it takes when the speed is doubled remain as

T /2 = 0.29 s

7 0

3 years ago

a 25 newton force applied on an object moves it 50 meters. the angle between the force and displacement is 40.0°. what is the va

LiRa [457]

Work = force × distance × cos(angle)

work = (25)(50)(cos (40))

work = 957.56 Joules

= 9.6x10^2 Joules

3 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 ball is thrown directly downward with an initial speed of 7.65 m/s froma height of 29.0 m. After what time interval does it st

coldgirl [10]

Answer: 1.77 s

Explanation: In order to solve this problem we have to use the kinematic equation for the position, so we have:

xf= xo+vo*t+(g*t^2)/2 we can consider the origin on the top so the xo=0 and xf=29 m; then

(g*t^2)/2+vo*t-xf=0 vo is the initail velocity, vo=7.65 m/s

then by solving the quadratric equation in t

t=1.77 s

8 0

3 years ago

How is it technically correct to say that a car making a u-turn can have a constant speed but cannot have a constant velocity?

saw5 [17]

During the "U" part of the turn, the car would follow an approximately circular path, and if it's moving at a constant speed, it would have to accelerate toward the center of the circle in order to change its direction.

5 0

2 years ago