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yawa3891 [41]
3 years ago
11

A block of mass m=1kg sliding along a rough horizontal surface is traveling at a speed v0=2m/s when it strikes a massless spring

head-on and compresses the spring a maximum distance X=10cm. If the spring has stiffness constant k=10N/m, determine the coefficient of kinetic friction between block and surface.
Physics
1 answer:
topjm [15]3 years ago
4 0

Here we can use the work energy theorem

W_f + W_s = K_f - K_i

here we know that

K_f = 0

as it come to rest finally

K_i = \frac{1}{2}mv_i^2

K_i = \frac{1}{2}\times 1\times 2^2

K_i = 2 J

now work done by friction force will be given as

W_f = - F_f \times d = -\mu mg d

W_f = - \mu(1)(9.8)(0.10) = - 0.98\mu

Work done by spring force is given as

W_s = \frac{1}{2}k(x_i^2 - x_f^2)

W_s = \frac{1}{2}(10)( 0 - 0.10^2)

W_s = -0.05 J

so now plug in all data above

- 0.05 - \mu(0.98) = 0 - 2

\mu = 1.99

so above is the friction coefficient


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List several examples of applied force, normal force, and friction that you’ve observed in your life.
grandymaker [24]

applied forces would be push for example.

normal forces would seem to be a force such as gravity.

friction for example when you try to slide on carpet but the fabric or whatever its made of stops you.

3 0
3 years ago
A ball is thrown straight up. What will
Papessa [141]

Answer:

velocity at the top: 0 m/s

acceleration at the top: -9.8 m/s²

Explanation:

Assuming up is positive and down is negative;

The velocity of the ball at the top of its path will be 0 m/s and the acceleration will be negative.

The velocity is 0 m/s because the ball does not move at the top of its path, and it switches from a positive velocity to a negative velocity. It must go through 0 in order to go from positive to negative.

The acceleration, however, is always negative no matter where the ball is in its motion. This negative acceleration causes the ball to slow down as it reaches the top, and speed up as it reaches the bottom.

<u>Think about it:</u> If there wasn't a negative acceleration, and it was instead 0, the ball would never come back down and instead keep going in a straight line.

6 0
2 years ago
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Which describes sublimation?
balu736 [363]
A. Solid turns into a gas
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2 years ago
The compressor on an air conditioner draws 35.0 A when it starts up. If the start-up time is 0.59 s, how much charge passes a cr
Elden [556K]

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is a conditioner es 13.4

6 0
3 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
3 years ago
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