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A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15āˆ˜ incline at a steady 15 m/s when, unfortunately, the poorly secured load falls off! Immediately after losing the load, the truck begins to accelerate at 1.5 m/s2. Part A What was the mass of the load? Ignore rolling friction.

Respuesta :

Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

Mass of the truck = 7100 kg

Angle = 15Ā°

velocity = 15m/s

Acceleration = 1.5 m/sĀ²

Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

ā‡’ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinĪø

ā‡’ After the load falls off, thrust (T) remains the same but downhill component of weight becomes Ā m1*gsinĪø .

Resultant force on truck is F = T ā€“ m1*gsinĪø Ā 

F causes the acceleration of the truck: F= m*a

This gives the equation:

T ā€“ m1*gsinĪø = m1*a Ā 

T = m1(a + gsinĪø)

Combining both equations gives:

(m1+m2)*g*sinĪø = m1*(a + gsinĪø)

m1*g*sinĪø + m2*g*sinĪø =m1*a + m1*g*sinĪø

m2*g*sinĪø = m1*a

Since m1+m2 = 7100kg, m1= 7100 ā€“ m2. This we can plug into the previous equation:

m2*g*sinĪø = (7100 ā€“ m2)*a

m2*g*sinĪø = 7100a ā€“ m2a

m2*gsinĪø + m2*a = 7100a

m2* (gsinĪø + a) = 7100a

m2 = 7100a/(gsinĪø Ā + a)

m2 = (7100 * 1.5) / (9.8sin(15Ā°) + 1.5)

m2 = 2636.8 kg

The load has a mass of 2636.8 kg

The mass of the load is 2636.8 kg

given information:

Mass of the truck m = 7100 kg

Angle Īø = 15Ā°

velocity v = 15m/s

Acceleration a = 1.5 m/sĀ²

Now let us assume that the mass of the truck is mā‚ and the mass of the load is mā‚‚. The thrust from the engine be T.

resolving forces:

Initially, the thrust (T) balances the total weight:

[tex]T = (m_1+m_2) g sin\theta[/tex]

When the load falls off, the thrust (T) remains the same

so the net force on the truck

[tex]F = T -m_1gsin\theta[/tex]

[tex]T-m_1gsin\theta = m_1a \\\\ T = m_1(a + gsin\theta)[/tex]

comparing both the equation for T we get:

[tex](m_1+m_2)gsin\theta = m_1(a + gsin\theta)\\\\m_1gsin\theta + m_2gsin\theta =m_1a + m_1gsin\theta\\\\m_2gsin\theta = m_1a[/tex]

Now, mā‚+mā‚‚ = 7100kg

mā‚= 7100 ā€“ mā‚‚

[tex]m_2gsin\theta = (7100-m_2)a\\\\m_2gsin\theta = 7100a-m_2a\\\\m_2gsin\theta + m_2a = 7100a\\\\m_2 (gsin\theta + a) = 7100a\\\\m_2 = 7100a/(gsin\theta + a)\\\\m_2 = (7100 \times1.5) / (9.8sin(15) + 1.5)\\\\m_2 = 2636.8 kg[/tex]

The load has a mass of 2636.8 kg

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