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Oksana_A [137]
2 years ago
12

A box with mass 15.0 kg moves on a ramp that is inclined at an angle of 55.0∘ above the horizontal. The coefficient of kinetic f

riction between the box and the ramp surface is μk = 0.300.
a. Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 190.0 N that is parallel to the ramp surface and directed down the ramp, moving the box down the ramp.
Express your answer with the appropriate units.
b. Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 190.0 N that is parallel to the ramp surface and directed up the ramp, moving the box up the ramp.
Express your answer with the appropriate units.
Physics
1 answer:
Nonamiya [84]2 years ago
8 0

Answer:

a) a = 19.0 m/s²

b) a = 2.9 m/s²

Explanation:

a) We draw the free body diagram of the box. There are 4 forces: the normal force N, the weight mg, the constant force F and the kinetic frictional force μ_kN. We can take the coordinate system which is rotated 55° from the horizontal, to ease the calculations. So, we write the equations of motion in each axis:

x: F+mg\sin\theta-\mu_k N=ma\\y: N-mg\cos\theta=0 \implies N=mg\cos\theta\\

Substituting the expression for N in the first equation, we have:

F+mg\sin\theta-\mu_kmg\cos\theta=ma\\\\\implies a=\frac{F+mg\sin\theta-\mu_kmg\cos\theta}{m}

If we plug in the given values, we have:

a=\frac{190.0N+(15.0kg)(9.8m/s^{2})\sin55\°-0.300(15.0kg)(9.8m/s^{2})\cos55\° }{15.0kg} =19.0m/s^{2}

Since we chose the right-downward direction as positive, the positive sign in this case means that the box is accelerated downwards above the ramp.

b) In this case, the constant force F and the kinetic frictional force μ_kN point to the opposite side. In other words, we can just only change the sign of this two forces in the equations of part (a) and obtain:

a=\frac{-F+mg\sin\theta+\mu_kmg\cos\theta}{m}

Plugging in the given values:

a=\frac{-190.0N+(15.0kg)(9.8m/s^{2})\sin55\°+0.300(15.0kg)(9.8m/s^{2})\cos55\° }{15.0kg} =-2.9m/s^{2}

Since we chose the right-downward direction as positive, the negative sign in this case means that the box is accelerated upwards above the ramp.

This means that the magnitude of the acceleration in this case is 2.9m/s².

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Answer:

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It is given that,

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The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

m_1u_1+m_2u_2=(m_1+m_2)V

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|J| = 4 kg-m/s

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a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres
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Answer:

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Part b) When collision is perfectly elastic

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Explanation:

Part a)

As we know that collision is perfectly inelastic

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mv_m = (m + M)v

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v = \frac{mv_m}{m + M}

now we know that in order to complete the circle we will have

v = \sqrt{5Rg}

\frac{mv_m}{m + M} = \sqrt{5Rg}

now we have

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b)

Now we know that collision is perfectly elastic

so we will have

v = \frac{2mv_m}{m + M}

now we have

\sqrt{5Rg} = \frac{2mv_m}{m + M}

v_m = \frac{m + M}{2m}\sqrt{5Rg}

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Since the electric field E is derived from the Coulomb Force per unit charge using a positive test charge, the field's units will be in units of Newtons/Coulomb, and be the formula for the Coulomb electric force between to charges (Q1 and Q2),

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Read 2 more answers
A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0∘above the horizontal by a rope exerting a 72.0-N pull
Elden [556K]

Answer:

(A) 374.4 J

(B) -332.8 J

(C) 0 J

(D) 41.6 J

(E)  351.8 J

Explanation:

weight of carton (w) = 128 N

angle of inclination (θ) = 30 degrees

force (f) = 72 N

distance (s) = 5.2 m

(A) calculate the work done by the rope

  • work done = force x distance x cos θ
  • since the rope is parallel to the ramp the angle between the rope and

        the ramp θ will be 0

       work done = 72 x 5.2 x cos 0

       work done by the rope = 374.4 J

(B) calculate the work done by gravity

  • the work done by gravity = weight of carton x distance x cos θ
  • The weight of the carton = force exerted by the mass of the carton = m x g
  • the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.

      work done by gravity = 128 x 5.2  x cos 120

      work done by gravity = -332.8 J

(C) find the work done by the normal force acting on the ramp

  • work done by the normal force = force x distance x cos θ
  • the angle between the normal force and the ramp is 90 degrees

       

         work done by the normal force = Fn x distance x cos θ

         work done by the normal force = Fn x 5.2 x cos 90

         work done by the normal force = Fn x 5.2 x 0

         work done by the normal force = 0 J

(D)  what is the net work done ?

  • The net work done is the addition of the work done by the rope,       gravitational force and the normal force

     net work done = 374.4 - 332.8 + 0 =  41.6 J  

(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal

  • work done by the rope= force x distance x cos θ
  • the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp

work done = 72 x 5.2 x cos 20

work done = 351.8 J

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2 years ago
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