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SSSSS [86.1K]
3 years ago
13

A 15,000 N truck starts from rest and moves down a 15∘ hill with the engine providing a 8,000 N force in the direction of the mo

tion. Assume the rolling friction force between the truck and the road is very small. If the hill is 50 m long, what will be the speed of the truck at the bottom of the hill?
Physics
1 answer:
GenaCL600 [577]3 years ago
7 0

Answer:

v_f = 27.9 m/s

Explanation:

Component of the weight of the truck along the inclined plane is given as

F_1 = W sin\theta

F_1 = 15000 sin15

F_1 = 3882.3 N

also the engine is providing the constant force to it as

F_2 = 8000 N

now the net force along the the plane is given as

F_{net} = 8000 + 3882.3

F = 11882.3 N

mass of the truck is given as

m = \frac{w}{g} = 1529 kg

now the acceleration is given as

a = \frac{F}{m}

a = 7.77 m/s^2

now the speed of the truck after travelling distance of d = 50 m is given as

v_f^2 = v_i^2 + 2 a d

v_f^2 = 0 + 2(7.77)(50)

v_f = 27.9 m/s

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3 years ago
After driving a portion of the route, the taptap is fully loaded with a total of 27 people including the driver, with an average
lara [203]

Answer:

compression of spring is x = 0.12 m

Assumed k = 160,000 N/m ........ Truck's suspension system

Explanation:

Given:

- The mass of average person m_p = 69 kg

- Total number of persons n_p = 27

- The mass of each goat m_g = 15 kg

- The total number of goats n_g = 3

- The mass of each chicken m_c = 3 kg

- The total number of goats n_c = 5

- The total mass of bananas m_b = 25 kg

Find:

How much are the springs compressed?

Solution:

- Using equilibrium equation on the taptap in vertical direction:

                                 F_net = F_spring - F_weight = 0

- Compute the force due to all the weights on the taptap:

                                F_weight = (n_p*m_p + n_g*m_g + n_c*m_c + m_b)*9.81

                                F_weight = (69*27 + 3*15 + 5*3 + 25)*9.81  

                                F_weight = 19109.88 N

- The restoring force of a spring is given by:

                                F_spring = k*x

Where, k is the spring stiffness and x is the displacement:

                                 F_weight = F_spring

                                 19109.88 = k*x

                                 x = 19109.88 / k

We need to assume the spring stiffness we will take k = 160,0000 N/m (trucks suspension systems). The value of the stiffness must be high enough to sustain a load of 1.911 tonnes.

                                 x = 19109.88 / 160,000

                                 x = 0.1194 m ≈ 0.12 m = 12 cm

- A compression of 12 cm seems reasonable for a taptap to carry 1.911 tonnes of load. Hence, the assumption of spring stiffness was reasonable. Hence, the compression of spring is x = 0.12 m.

8 0
3 years ago
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