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VLD [36.1K]
3 years ago
14

You place a box weighing 200.2 N on a inclined plane that makes a 37.1 angle with the horizontal. Compute the component of the g

ravitational force acting down on the inclined plane. Answer in Units of N

Physics
1 answer:
salantis [7]3 years ago
8 0

Answer:

141.56 N.

Explanation:

Data given:

Weight of the box= 200.2 N

Angle with the horizontal= 37.1°

Solution;

Gravitational force on the box, F_g= weight of the box

                                                                           = 200.2 N

Component of gravitational force along plane = F_g * sin( ∅ )

                                                                                   = W * (sin∅)

                                                                                   = (200.1) * sin (37.1°)

                                                                                   = 141.56 N

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Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is   spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 .  t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

    p₀ = m1 v₀₁

Moment after shock

    p_{f} = (m1 + m2) v_{f}

   p₀ = p_{f}

   m1 v₀₁ = (m1 + m2) v_{f}

  v_{f} = v₀₁ m1 / (m1 + m2)

   v_{f}= 4.4 600 / (600 + 500)

  v_{f} = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

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   Em₀ = K = ½ (m1 + m2) v_{f}²

After compressing the spring

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As there is no rubbing the energy is conserved

   Em₀ = E_{mf}

   ½ (m1 + m2) v_{f}² = = ½ k x²

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   x = 2.4 √ (11/3000)

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A student jumps off a sled toward the NORTH after it stops at the bottom of an icy hill. Based on Newton's third law of motion,
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When the student the sled jumps off toward the north , the sled most likely move towards the south.

<h3>What is the Newton third law?</h3>

According to the Newton third law of motion, action and reaction are equal and opposite. This means that the direction of the reaction force must also be opposite to that of the action.

As such, when the student the sled jumps off toward the north , the sled most likely move towards the south.

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A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D
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Reemplazando:

(0 m/s)² - (50 m/s)² = 2*a*120 m

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