Question

A dump truck moves up an incline of 5° with constant tractive force of 800 kN. The gross mass of the truck is 250 tonne and its rolling resistance is 545 kN. The acceleration due to gravity is 10 m²/s. The time required, in s, to reach a speed of 3.3 m/s from 1.0 m/s is

• A. 22.0
• B. 15.5
• C. 3.3
• D. 0.2

Hint:

Always check for forces acting in the direction of motion (tractive force, gravity) and calculate the net force to determine acceleration.

Answer 1

The Correct Option is B

Step 1: Identify forces acting on the truck.
Forces acting on the truck include:
- Tractive force: 800 kN
- Rolling resistance: 545 kN
- Component of the gravitational force along the incline: \( \text{Force due to gravity} = m \cdot g \cdot \sin(\theta) \)
Where:
- \( m = 250 \, \text{tonne} = 250 \times 10^3 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( \theta = 5^\circ \)

Step 2: Net force calculation.
The net force available to accelerate the truck is: \[ F_{\text{net}} = \text{Tractive force} - \text{Rolling resistance} - \text{Component of gravity along incline} \] The component of gravity along the incline: \[ F_{\text{gravity}} = m \cdot g \cdot \sin(\theta) = 250 \times 10^3 \cdot 10 \cdot \sin(5^\circ) = 250 \times 10^3 \cdot 10 \cdot 0.087 = 217500 \, \text{N}. \] So, the net force is: \[ F_{\text{net}} = 800000 - 545000 - 217500 = 375000 \, \text{N}. \]

Step 3: Apply Newton's second law.
From Newton's second law, \( F_{\text{net}} = m \cdot a \), so we can solve for acceleration \( a \): \[ a = \frac{F_{\text{net}}}{m} = \frac{375000}{250 \times 10^3} = 1.5 \, \text{m/s}^2. \]

Step 4: Use the kinematic equation to find the time.
The kinematic equation is: \[ v = u + at, \] where:
- \( v = 3.3 \, \text{m/s} \) (final speed)
- \( u = 1.0 \, \text{m/s} \) (initial speed)
- \( a = 1.5 \, \text{m/s}^2 \) (acceleration)
Rearranging for time \( t \): \[ t = \frac{v - u}{a} = \frac{3.3 - 1.0}{1.5} = \frac{2.3}{1.5} = 15.5 \, \text{s}. \]

Final Answer: 15.5 seconds.