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Scrat [10]
3 years ago
10

Hint: This problem requires a train of logic. (1) Analyze force diagram, (2) use Newton’s Laws, and (3) solve the equations of m

otion. A block starts from rest at a height of 3.9 m on a fixed inclined plane. The acceleration of gravity is 9.8 m/s 2 . 3.8 kg µ = 0.12 31◦ What is the speed of the block at the bottom of the ramp? Answer in units of m/s.
Physics
1 answer:
gregori [183]3 years ago
6 0

8 . 55599 m / s.  
Let : h = 8 . 2 m , m = 5 . 1 kg , = 0 . 22 , = 22 , and v f = final speed . The normal force to the inclined plane is N = mg cos . The sum of the forces par- allel to the inclined plane is F net = ma = mg sin - mg cos a = g sin - g cos Since v 2 f = v 2 + 2 ax = 2 ad (1) along the plane, and neglecting the dimension of the block, the distance, d, to the end of the ramp is d sin = h d = h sin (2) therefore v f = r 2 ah sin = r 2 g h (sin - cos ) sin = p 2 g h (1- cot ) (3) = q 2(9 . 8 m / s 2 )(8 . 2 m)[1- (0 . 22)cot22 ] = 8 . 55599 m / s .
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How far will 350 j raise a 7 kg mass?
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6 0
3 years ago
a 25-N net force is applied to a rolling cart and pruduces an acceleration of 5 m/s 2 what’s the cart mass
nignag [31]

Answer:

The mass of the cart is 5 kg

Explanation:

You divide 25 by 5 and get 5. Have a great day! :D

<em>The Equation:</em>

25/5 = 5

8 0
2 years ago
in a lever, a load of 600N is lifted by using 400 effort. If the load is at the distance of 20cm and the effort at the distance
koban [17]

Explanation:

Load (l) = 680N

Effort (E) = 500N

Length slope (l) = 12m

Height slope (h) = 8 m

Output = load * height

680 *8 = 5.44 *103 J

The Input = effort * length = 500 *12 = 6000J

the Mechanical advantage (M.A) = load effort= 600500=1.36

the Velocity ratio (V.R) =lh=128 = 1.5

the Efficiency =M.A100%V.R= 90.6%

8 0
2 years ago
A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic
Ilia_Sergeevich [38]

Answer:

\mu_k=0.27

Explanation:

According to the free body diagram, in this case, we have:

\sum F_x:-F_k=ma\\\sum F_y:N=mg

Recall that the force of friction is given by:

F_k=\mu_k N

Replacing and solving for the coefficient of kinetic friction:

-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}

Finally, we calculate \mu_k:

\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27

4 0
3 years ago
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