A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ 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.

Answer :

Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

Mass of the truck = 7100 kg

Angle = 15°

velocity = 15m/s

Acceleration = 1.5 m/s²

Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinθ

⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes  m1*gsinθ .

Resultant force on truck is F = T – m1*gsinθ  

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

This gives the equation:

T – m1*gsinθ = m1*a  

T = m1(a + gsinθ)

Combining both equations gives:

(m1+m2)*g*sinθ = m1*(a + gsinθ)

m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ

m2*g*sinθ = m1*a

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

m2*g*sinθ = (7100 – m2)*a

m2*g*sinθ = 7100a – m2a

m2*gsinθ + m2*a = 7100a

m2* (gsinθ + a) = 7100a

m2 = 7100a/(gsinθ  + a)

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

velocity v = 15m/s

Acceleration a = 1.5 m/s²

Now let us assume that the mass of the truck is m₁ and the mass of the load is m₂. 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, m₁+m₂ = 7100kg

m₁= 7100 – m₂

[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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