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Question
3 formula 1 point a box has a mass of 68 kg and sits on a flat floor at rest. determine the magnitude of the normal force that acts on the box. answer 4 formula 1 point a box has a mass of 7 kg and sits on a flat floor. the box has a coefficient of static friction equal to 0.7. what is the greatest force that can be applied to the object before it starts to move? answer 5 multiple choice 1 point a heavy box can remain at rest while being pushed with an applied force of up to 900 newtons. if the box is pushed with a force slightly larger than this, what will happen? the box will start to move and the kinetic friction experienced by the box will be more than 900 newtons the box will remain at rest and the static friction force will be equal to 900 newtons. the box will start to move and the kinetic friction experienced by the box will be less than 900 newtons the box will remain at rest and the static friction force will be equal to 901 newtons. the box will start to move and the kinetic friction experienced by the box will be equal to 900 newtons
Step1: Calculate the normal force for question 3
According to Newton's second law, when the box is at rest on a flat floor, the normal force \(N\) equals the weight \(mg\). Given \(m = 68\space kg\) and \(g=9.8\space m/s^{2}\), we have \(N = mg\).
Step2: Calculate the maximum static - friction force for question 4
The formula for the maximum static - friction force \(F_{s,max}=\mu_{s}N\). Since \(N = mg\) (where \(m = 7\space kg\) and \(g = 9.8\space m/s^{2}\)), and \(\mu_{s}=0.7\). First, calculate \(N=7\times9.8 = 68.6\space N\). Then \(F_{s,max}=\mu_{s}N\).
Step3: Analyze question 5
When the applied force is slightly larger than the maximum static - friction force, the box starts to move. The kinetic - friction force \(F_{k}=\mu_{k}N\). Since \(\mu_{k}<\mu_{s}\), \(F_{k}
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- The box will start to move and the kinetic friction experienced by the box will be less than 900 Newtons.