How Is the Third Answer Calculated in This Lawn Mower Physics Problem?

  • Thread starter Thread starter justinbaker
  • Start date Start date
AI Thread Summary
The third answer in the lawn mower physics problem is calculated using Newton's Second Law, where the required force to accelerate the mower is determined by the mass and acceleration. The mass of the mower is 17.0 kg, and the desired acceleration is 1.1 m/s², resulting in a force of 18.7 N. To account for the existing retarding force of 49.3 N, the total force needed becomes 68.0 N. Given that the person is pushing with a force of 71.0 N, the effective force for acceleration is 21.7 N. This calculation clarifies how to arrive at the required force for the mower's acceleration.
justinbaker
Messages
34
Reaction score
0
Here is the question i have. I got the first two no prob. But ihave no idea how i got the third answer. The answers shown are all correct. So if anyone can explain how to get the third answer, thanks everyone

A person pushes a 17.0 kg lawn mower at constant speed with a force of 71.0 N directed along the handle, which is at an angle of = 46.0° to the horizontal

http://www.webassign.net/giancoli5/4_40alt.gif

Calculate the horizontal retarding force on the mower = 49.3 N

Calculate the normal force exerted vertically upward on the mower by the ground =218

Calculate the force the person must exert on the lawn mower to accelerate it from rest to 1.1 m/s in 2.0 seconds (assuming the same retarding force).
=82.8


Thanks,
justinbaker@gmail.com
 
Physics news on Phys.org
Since the mover moves in the horizontal direction only, we will consider the x-components only. Left direction : Negative; Right direction: Positive.
<br /> \begin{multline*}<br /> \begin{split}<br /> &amp;Newton&#039;s\ 2nd\ Law:\\<br /> &amp;\sum \vec{F}=m\vec{a}\\<br /> &amp;\vec{F_x}+\vec{D}=m\vec{a}\\<br /> &amp;|\vec{F}|cos46^0 + (-49.3)=17.0*(\frac{1.1-0}{2.0})\\<br /> &amp;|\vec{F}|=84.4\ N\\<br /> \end{split}<br /> \end{multline*}<br />
 

Attachments

  • Picture.GIF
    Picture.GIF
    963 bytes · Views: 517


The third answer, 82.8 N, can be calculated using Newton's Second Law of Motion, which states that force equals mass times acceleration (F=ma). In this scenario, the mass of the lawn mower is given as 17.0 kg and the acceleration is 1.1 m/s^2. Therefore, the force needed to accelerate the mower can be calculated as 17.0 kg x 1.1 m/s^2, which equals 18.7 kg*m/s^2 or 18.7 N. However, this only accounts for the force needed to overcome the retarding force (49.3 N) and does not take into account the force needed to maintain the constant speed of the mower. So, the total force needed is 49.3 N + 18.7 N, which equals 68.0 N. However, the problem states that the person is pushing with a force of 71.0 N, so the actual force needed to accelerate the mower would be 71.0 N - 49.3 N, which equals 21.7 N. This is the force that the person must exert on the lawn mower to accelerate it from rest to 1.1 m/s in 2.0 seconds. I hope this helps clarify the calculation for you.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top