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zepelin [54]
3 years ago
5

The two blocks are connected by a light string that passes over a frictionless pulley with a negligible mass. The 2 kg block lie

s on a rough horizontal surface with a constant coefficient of kinetic friction 0.4. This block is connected to a spring with spring constant 4 N/m. The second block has a mass of 8 kg. The system is released from rest when the spring is unstretched, and the 8 kg block falls a distance h before it reaches the lowest point. The acceleration of gravity is 9.8 m/s2 . Note: When the 8 kg block is at the lowest point, its velocity is zero.
a. Calculate the falling distance h where the 8 kg blocks stops.
b. Consider the moment when the 8 kg block has descended by a distance of 21.168 m, where 21.168 m is less than h. At this moment calculate the sum of the kinetic energy for the two blocks. Answer in units of J.
Physics
1 answer:
frozen [14]3 years ago
5 0

Answer: YOU ARE A FAT NEGRO

Explanation:YOU ARE

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Explanation:

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5 0
3 years ago
A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk
Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

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The car acceleration is equal to the acceleration of the truck

ac: car acceleration\frac{m}{s^{2} }

at: truck acceleration\frac{m}{s^{2} })

ac = at= \frac{vf-vi}{t-ti}  equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 \frac{m}{s}, t = final time =5 s

We replaced the known information in the equation(1):

ac = at = \frac{13-0}{15-0}

ac=ac=\frac{13}{15}  \frac{m}{s}

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

Wc=mc*g

mc = Car mass = 2230kg

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Wc= 2230*9.8

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Normal force calculation:

Newton's first law

sum Fy= 0

N-W=0

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N=21854 N

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N  Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

Ff=0.373*21854

Ff=8151.54 N

Calculation of the tension force in the rope (T):

Newton's Second law

sum Fx= mc*ac

T-Ff=mc*ac

T=2230(\frac{13}{15}) + 8151.54

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0
3 years ago
A person is sitting at the very back of a canoe of length L, when the front just bumps into the dock. show answer No Attempt 50%
Pavel [41]

The distance of the canoeist from the dock is equal to length of the canoe, L.

<h3>Conservation of linear momentum</h3>

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where;

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Answer:

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Given: t = 2.0 s, g = 10 m/s², u = 0 m/s (dropped from height)

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Answer:

Impulse will be 12 kgm/sec

So option (b) will be correct option

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So option (b) will be correct option

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3 years ago
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