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Math  /  Algebra

Questionastronaut, the vehicle in which she orbits experiences an equal and opposite force. Use this knowledge to find an equation for the acceleration of the system (astronaut and spaceship) that would be measured by a nearby observer. (c) Discuss how this would affect the measurement of the astronaut's acceleration. Propose a method by which recoil of the vehicle is avoided.
28. In Figure 5.4.3, the net external force on the 24−kg24-\mathrm{kg} mower is given as 51 N . If the force of friction opposing the motion is 24 N , what force F '(in newtons is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s1.5 \mathrm{~m} / \mathrm{s} when the force FF is removed. How far will the mower go before stopping?

Studdy Solution

STEP 1

1. The net external force on the mower is given as 51 N.
2. The force of friction opposing the motion is 24 N.
3. The mass of the mower is 24 kg.
4. The initial velocity of the mower is 1.5 m/s when the force is removed.
5. We need to find the force F′ F' exerted by the person and the stopping distance of the mower.

STEP 2

1. Determine the force exerted by the person on the mower.
2. Calculate the acceleration of the mower when the force is removed.
3. Determine the stopping distance of the mower.

STEP 3

To find the force F′ F' exerted by the person, use the equation for net force: Fnet=F′−Ffriction F_{\text{net}} = F' - F_{\text{friction}}
Given: Fnet=51 N F_{\text{net}} = 51 \, \text{N} Ffriction=24 N F_{\text{friction}} = 24 \, \text{N}
Substitute the known values: 51=F′−24 51 = F' - 24
Solve for F′ F' : F′=51+24=75 N F' = 51 + 24 = 75 \, \text{N}

STEP 4

To find the acceleration when the force is removed, use Newton's second law: Fnet=m⋅a F_{\text{net}} = m \cdot a
The only force acting is friction, so: Ffriction=m⋅a F_{\text{friction}} = m \cdot a
Substitute the known values: 24=24⋅a 24 = 24 \cdot a
Solve for a a : a=2424=1 m/s2 a = \frac{24}{24} = 1 \, \text{m/s}^2
The acceleration is negative because it opposes the motion: a=−1 m/s2 a = -1 \, \text{m/s}^2

STEP 5

To find the stopping distance, use the kinematic equation: v2=u2+2as v^2 = u^2 + 2as
Where: - v=0 m/s v = 0 \, \text{m/s} (final velocity) - u=1.5 m/s u = 1.5 \, \text{m/s} (initial velocity) - a=−1 m/s2 a = -1 \, \text{m/s}^2 (acceleration) - s s is the stopping distance
Substitute the known values: 0=(1.5)2+2(−1)s 0 = (1.5)^2 + 2(-1)s
Solve for s s : 0=2.25−2s 0 = 2.25 - 2s 2s=2.25 2s = 2.25 s=2.252=1.125 m s = \frac{2.25}{2} = 1.125 \, \text{m}
The force exerted by the person is 75 N 75 \, \text{N} and the stopping distance is 1.125 m 1.125 \, \text{m} .

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