Question

Difficulty: MediumMeasurement of Mass and Weight

A rocket carries a payload of mass 25 kg25\text{ kg} resting on a spring balance. During its vertical ascent, the spring balance indicates a reading of 350 N350\text{ N}. Taking the acceleration due to gravity as g=10 m s2g = 10\text{ m s}^{-2}, what is the magnitude of the upward acceleration of the rocket in m s2\text{m s}^{-2}?

Answer: 4 m s^-2

Answer

The upward acceleration of the rocket is 4.0 m s24.0\text{ m s}^{-2}.
When an object of mass mm accelerates upward at rate aa, the scale must exert an upward force RR that overcomes the weight mgmg and provides net acceleration mama, such that R=m(g+a)R = m(g + a). Substituting R=350 NR = 350\text{ N}, m=25 kgm = 25\text{ kg}, and g=10 m s2g = 10\text{ m s}^{-2} yields 350=25(10+a)350 = 25(10 + a), which simplifies to a=4.0 m s2a = 4.0\text{ m s}^{-2}.

Step-by-Step Solution

1
Formulate the equation of motion for apparent weight during upward acceleration
Rmg=ma    R=m(g+a)R - mg = ma \implies R = m(g + a)
The spring balance supports the mass against gravity while simultaneously providing the force for upward acceleration.
2
Substitute the given numerical parameters into the relation
350=25(10+a)350 = 25(10 + a)
The measured reading R=350 NR = 350\text{ N}, mass m=25 kgm = 25\text{ kg}, and standard gravity g=10 m s2g = 10\text{ m s}^{-2} are supplied.
3
Solve the algebraic equation for the rocket's acceleration aa
a=4.0 m s2a = 4.0\text{ m s}^{-2}
Dividing 350 N350\text{ N} by 25 kg25\text{ kg} gives an effective total acceleration of 14 m s214\text{ m s}^{-2}; subtracting g=10 m s2g = 10\text{ m s}^{-2} isolates the upward acceleration aa.

Key Concept

Apparent Weight and Mass Measurement in Accelerating Frames
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