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Difficulty: MediumMechanics and Motion

A conveyor system accelerates a 15 kg15\text{ kg} package along a straight horizontal track from an initial speed of 2 m/s2\text{ m/s} to a final speed of 8 m/s8\text{ m/s} in 3 seconds3\text{ seconds}. What is the magnitude of the net horizontal force acting on the package, in newtons?

Answer: 30 N

Answer

The magnitude of the net horizontal force acting on the package is 30 N30\text{ N}.
By finding the acceleration through a=vf−vit=8−23=2 m/s2a = \frac{v_f - v_i}{t} = \frac{8 - 2}{3} = 2\text{ m/s}^2 and applying Newton's second law (F=maF = ma), the resulting net force is 15 kg×2 m/s2=30 N15\text{ kg} \times 2\text{ m/s}^2 = 30\text{ N}.

Step-by-Step Solution

1
Determine the horizontal acceleration of the package using 1D kinematics.
a=8 m/s−2 m/s3 s=6 m/s3 s=2 m/s2a = \frac{8\text{ m/s} - 2\text{ m/s}}{3\text{ s}} = \frac{6\text{ m/s}}{3\text{ s}} = 2\text{ m/s}^2
Acceleration measures the rate of change of velocity over elapsed time (a=ΔvΔta = \frac{\Delta v}{\Delta t}).
2
Calculate the net force by applying Newton's second law of motion.
Fnet=m×a=15 kg×2 m/s2=30 NF_{\text{net}} = m \times a = 15\text{ kg} \times 2\text{ m/s}^2 = 30\text{ N}
Net force is directly proportional to the mass and the acceleration of the object (F=maF = ma).

Key Concept

Newton's Second Law and Uniform Acceleration
Estimated Time:1m 0s
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Topics

Mechanics and Motion

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