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4vir4ik [10]
3 years ago
12

The kinetic friction coefficient between a cabinet and the floor is 0.3. Mass of the cabinet is 300kg. A man pushes the cabinet

along the floor with a 1000N force. Find the acceleration of the cabinet. g
Physics
1 answer:
podryga [215]3 years ago
7 0

Answer:

<h2>0.39m/s^2</h2>

Explanation:

Step one:

given data

mass m= 300kg

applied force F= 1000N

coefficient of friction μ= 0.3

Step two:

The net force Fn= applied force-friction force  

Fn=F-F1

F1= limiting force

F1=μ*m*g

F1=0.3*300*9.81

F1=882.9N

the Net force= 1000-882.9

Fn=117.1N

Step three:

we know that

F=ma

Fnet=ma

a= Fnet/m

a=117.1/300

a=0.39m/s^2

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Instead, in this problem the voltage applied is doubled:

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Two balls with equal masses, m, and equal speed, v, engage in a head on elastic collision. what is the final velocity of each ba
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- Let's start with conservation of momentum. The initial momentum of the total system is the sum of the momenta of the two balls, but we should put a negative sign in front of the velocity of the second ball, because it travels in the opposite direction of ball 1. So ball 1 has mass m and speed v, while ball 2 has mass m and speed -v:
p_i = p_1-p_2 = mv-mv =0
So, the final momentum must be zero as well:
p_f = 0
Calling v1 and v2 the velocities of the two balls after the collision, the final momentum can be written as
p_f = mv_1 + mv_2 = 0
From which
v_1 = -v_2

- So now let's apply conservation of kinetic energy. The kinetic energy of each ball is \frac{1}{2} mv^2. Therefore, the total kinetic energy before the collision is
K_i = \frac{1}{2} mv^2 +  \frac{1}{2} mv^2 = mv^2
the kinetic energy after the collision must be conserved, and therefore must be equal to this value:
K_f = K_i = mv^2 (1)
But the final kinetic energy, Kf, is also
K_f =  \frac{1}{2} mv_1^2 +  \frac{1}{2}mv_2^2
Substituting v_1 = -v_2 as we found in the conservation of momentum, this becomes
K_f = mv_2 ^2
we also said that Kf must be equal to the initial kinetic energy (1), therefore we can write 
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This means that after the collision, the two balls have same velocity v, but they go in the opposite direction with respect to their original direction.

8 0
3 years ago
9. What is the total kinetic energy (KEtran + KErot) of a solid cylinder with mass M = 2.50 kg and radius R = 0.5 m that rolls w
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Answer:

110.25 J

Explanation:

We are given that

Mass,M=2.5 kg

Radius,R=0.5 m

Distance,d=4.5 m

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According to law of conservation of energy

Total kinetic energy=Potential energy=mgh

Where g=9.8m/s^2

Using the formula

Total kinetic energy=2.5\times 9.8\times 4.5

Total kinetic energy=110.25 J

4 0
3 years ago
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