Question

10 formula 1 point a snail pushes against a kernel of corn and moves the kernel of corn at a constant velocity. if the kernel of corn experiences a kinetic friction force of 2.0 newtons, what must be the magnitude of the force applied by the snail? answer 11 formula 1 point a student applies a force of 122 newtons to a box to the right. the box has a mass of 7 kg and experiences a kinetic friction force to the left. the box has an acceleration of 4.7 m/s². determine the coefficient of kinetic friction for the box. answer

Explanation:

Step1: Analyze the motion of the kernel of corn

Since the kernel of corn moves at a constant velocity, the net force acting on it is zero. According to Newton's second law \(F_{net}=ma\), and \(a = 0\) (constant velocity), so \(F_{net}=0\).
The forces acting on the kernel are the force applied by the snail \(F_{snail}\) and the kinetic - friction force \(F_f\). So \(F_{net}=F_{snail}-F_f = 0\).

Step2: Solve for the force applied by the snail

From \(F_{snail}-F_f = 0\), we can get \(F_{snail}=F_f\). Given \(F_f = 2.0\space N\), so \(F_{snail}=2.0\space N\).

Step3: Analyze the motion of the box

For the box, according to Newton's second law \(F_{net}=F - F_f=ma\). The normal force \(N = mg\) (assuming the surface is horizontal), and the kinetic - friction force \(F_f=\mu_kN=\mu_kmg\). So \(F-\mu_kmg=ma\).

Step4: Solve for the coefficient of kinetic friction \(\mu_k\)

We can re - arrange the equation \(F-\mu_kmg=ma\) to \(\mu_k=\frac{F - ma}{mg}\).
Given \(F = 122\space N\), \(m = 7\space kg\), \(a=4.7\space m/s^2\), and \(g = 9.8\space m/s^2\).
First, calculate \(ma=7\times4.7 = 32.9\space N\) and \(mg=7\times9.8=68.6\space N\).
Then \(F - ma=122-32.9 = 89.1\space N\).
So \(\mu_k=\frac{89.1}{68.6}\approx1.3\).

Answer:

1. \(2.0\space N\)
2. \(1.3\)