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RUDIKE [14]
3 years ago
14

horizontal force pushes block up a 20.0 incline with an initial speed 12.0 m//s. a) how high up the plane does slide before comi

ng to rest? b)how much time does it take to return to its starting position?
Physics
1 answer:
Annette [7]3 years ago
7 0

Answer:

Explanation:

We shall apply conservation of mechanical energy .

initial kinetic energy = 1/2 m v²

= .5 x m x 12 x 12

= 72 m

This energy will be spent to store potential energy . if h be the height attained

potential energy = mgh , h is vertical height attined by block

= mg l sin20  where l is length up the inclined plane  

for conservation of mechanical energy

initial kinetic energy  = potential energy

72 m = mg l sin20

l = 72 /  g  sin20

= 21.5 m

deceleration on inclined plane = g sin20

= 3.35 m /s²

v = u - at

t = v - u / a

= (12 - 0) / 3.35

= 3.58 s

it will take the same time to come back . total time taken to reach original point = 2 x 3.58

= 7.16 s

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Answer: 1.3 Hz

Explanation:

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3 years ago
A large refrigerator (mass 80kg) sits at rest inside a house. The homeowner wants to move the refrigerator across the room, so s
Liono4ka [1.6K]

Answer:

400 N

Explanation:

By the law of friction,

F=\mu R

F is the maximum frictional force, \mu is the coefficient of friction and R is the reaction on the refrigerator. On a horizontal surface, the reaction is equal to the weight of the refrigerator.

R=mg

F=\mu mg

While not moving, the fricition on the refrigerator is static friction. So, \mu=0.65

F=0.65 \times80\times9.8=509.6 \text{ N}

This is the maximum frictional force and is more than the applied horizontal force of 400 N. Frictional force cannot be more than the applied force, else it would actually pull the refrigerator backwards (a strange thing, if it were to happen). It is equal to the extent of the applied force because the applied force is not enough to overcome the maximum.

Hence the frictional force is 400 N.

PS: Note that we do not use the coefficient of kinetic friction because applied force could not overcome the static friction.

6 0
3 years ago
How do you determine the speed or velocity of an object using a distance vs. time graph
Taya2010 [7]
The speed of an object can be determined from the distance vs time graph.

You know that speed = distance/time

in the graph, distance/time = slope of the curve.

So SPEED IS GIVEN BY THE SLOPE of the curve in the graph.

● If the distance vs time curve is a straight line, parallel to time axis(x-axis), slope is 0. That means speed is 0. So the object is at rest.

● If the distance vs time curve is a straight line, with some non-zero slope; That means speed is nonzero and constant. So the object is in uniform motion.

● If the distance vs time curve is a curved, the slope is changing. That means speed is changing. So the object is in an accelerated motion.
4 0
3 years ago
A 94 g particle undergoes SHM with an amplitude of 8.3 mm, a maximum acceleration of magnitude 7.8 x 103 m/s2, and an unknown ph
Lelechka [254]

Answer:

a) T = 6.49*10^-3 s

b) v = 8 m/s

c) E = 3 J

d) F = 733 N

e) F = 366.5 J

Explanation:

Given

Mass of particle, m = 94 g = 0.094 kg

Amplitude of the particle, A = 8.3 mm = 8.3*10^-3 m

Maximum acceleration of particle, a = 7.8*10^3 m/s²

the equation describing Simple Harmonic Motion is given as

x = A cos (wt +φ)

To fond the acceleration of this relationship, we would have to integrate. Twice, the first would be a Velocity, and the second acceleration that we need.

Velocity = dx/dt = -Aw sin(wt + φ)

Acceleration = d²x/dt = -Aw² cos(wt + φ)

From the question, we were given, magnitude of acceleration to be 7.8*10^3 m/s²

Aw² = 7.8*10^3

w² = 7.8*10^3 / A

w² = 7.8*10^3 / 8.3*10^-3

w² = 939759

w = √939759

w = 969

Recall, T = 2π/w, so that

T = (2 * 3.142) / 969

T = 6.49*10^-3 s

Maximum speed = Aw

Maximum speed = 8.3*10^-3 * 969

Maximum speed = 8.0 m/s

Total mechanical energy oscillator =

mgx + 1/2mx² =

1/2mv(max)² =

1/2 * 0.094 * 8² =

3 J

Maximum displacement

x = A cos(wt + φ)

For x to be maximum here, then cos(wt + φ) Must be equal to 1

Acceleration = d²x/dt² = -Aw²

And force = mass * acceleration

Force = 0.094 * 7.8*10^3

Force = 733 N

x = A cos(wt + φ), where cos(wt + φ) = 1/2

d²x/dt² = -Aw² * 1/2

d²x/dt² = 733 * 0.5

= 366.5 N

7 0
3 years ago
A 180 lb crate is on the ground, and a strong rope is attached. You need to move it across the basement floor, which has a coeff
jekas [21]

Answer:

F = 505.13 N

Yes it is better to pull the rope rather than push it

Explanation:

Let the force is applied at an angle of 60 degree

so we will have net vertical force on the crate is given as

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here we know

m = 180 lb

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F_n = 801 - 0.866 F

now friction force on the crate is given as

F_x = \mu F_n

Fcos60 = 0.7(801 - 0.866 F)

0.5F + 0.61F = 560.7

F = 505.13 N

Yes it is better to pull the rope rather than push it

6 0
3 years ago
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